Molrekenen voor een Nylon parachutedoek

Leerdoelen

•Je kunt uitleggen waarom het zinvol is om de hoeveelheid beginstof te kunnen uitrekenen.

•Je kunt een stappenplan toepassen om te rekenen aan reactievergelijkingen met behulp van mol.

Stap voor stap berekening

Stap A: Bereken het aantal mol ammoniak

1.Bepaal de massa in gram: 2000 kg = 2.000.000 gram

2.Bereken de molaire massa van ammoniak (NH_3): (stikstof) = 14\;g/mol14g/mol\frac{14}{\placeholder{}}g/mol (waterstof) = 1{,}01\;g/mol1{,}0\;g/mol1{,}\;g/mol1{,}o\;g/mol1{,}o\;g/mol1{,}\;g/mol1\;g/mol1{,}\;g/mol1{,}0\;g/mol1{,}01\;g/mol1{,}0\;g/mol1{,}\;g/mol1\;g/mol1m\;g/mol1\;g/mol1{,}\;g/mol1{,}0\;g/mol1{,}01\;g/mol1{,}0\;g/mol1{,}\;g/mol1\;g/mol Molaire massaNH_3NH_{\placeholder{}}is14+(3\times1{,}01)=17{,}03\;g/molN14+(3\times1{,}01)=17{,}03\;g/molNH14+(3\times1{,}01)=17{,}03\;g/molNH_{}14+(3\times1{,}01)=17{,}03\;g/molNH_314+(3\times1{,}01)=17{,}03\;g/molNH_3=14+(3\times1{,}01)=17{,}03\;g/molNH_3=14+(3\times1{,}01)=17{,}0\;g/molNH_3=14+(3\times1{,}01)=17{,}\;g/molNH_3=14+(3\times1{,}01)=17\;g/molNH_3=14+(3\times1{,}0)=17\;g/molNH_3=14+(3\times1{,})=17\;g/molNH_3=14+(3\times1)=17\;g/molNH_3=14+(3\times1)=17{,}\;g/molNH_3=14+(3\times1)=17{,}0\;g/molNH_3=14+(3\times1)=17{,}03\;g/molNH_3=14+(3\times1{,})=17{,}03\;g/molNH_3=14+(3\times1{,}0)=17{,}03\;g/molNH_3=14+(3\times1{,}01)=17{,}03\;g/molNH_3=14+(3\times1{,}01)=17{,}0\;g/molNH_3=14+(3\times1{,}01)=17{,}02\;g/molNH_3=14+(3\times1{,}01)=17{,}0\;g/molNH_3=14+(3\times1{,}01)=17{,}\;g/molNH_3=14+(3\times1{,}01)=17{,}{,}\;g/molNH_3=14+(3\times1{,}01)=17{,}{,}0\;g/molNH_3=14+(3\times1{,}01)=17{,}{,}\;g/molNH_3=14+(3\times1{,}01)=17{,}\;g/molNH_3=14+(3\times1{,}01)=17{,}0\;g/molNH_3=14+(3\times1{,}01)=17{,}01\;g/molNH_3=14+(3\times1{,}01)=17{,}0\;g/molNH_3=14+(3\times1{,}01)=17{,}\;g/molNH_3=14+(3\times1{,}01)=17\;g/molNH_3=14+(3\times1{,}0)=17\;g/molNH_3=14+(3\times1{,})=17\;g/molNH_3=14+(3\times1)=17\;g/molNH_3=14+(3\times1{,})=17\;g/molNH_3=14+(3\times1{,}0)=17\;g/molNH_3=14+(3\times1{,}01)=17\;g/molNH_3=14+(3\times1{,}0)=17\;g/molNH_3=14+(3\times1{,})=17\;g/molNH_3=14+(3\times1)=17\;g/molNH_3=14+(3\times1)=17g/molNH=14+(3\times1)=17g/mol

3.Bereken het aantal mol ammoniak: n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17{,}03\text{ g/mol}}\approx117439{,}81\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17{,}03\text{ g/mol}}\approx117439{,}8\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17{,}03\text{ g/mol}}\approx117439{,}\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17{,}03\text{ g/mol}}\approx117439{,}0\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17{,}03\text{ g/mol}}\approx117439{,}06\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17{,}03\text{ g/mol}}\approx11743947{,}06\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17{,}03\text{ g/mol}}\approx1174347{,}06\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17{,}03\text{ g/mol}}\approx117447{,}06\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17{,}03\text{ g/mol}}\approx11747{,}06\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17{,}03\text{ g/mol}}\approx117647{,}06\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17{,}0\text{ g/mol}}\approx117647{,}06\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17{,}\text{ g/mol}}\approx117647{,}06\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17\text{ g/mol}}\approx117647{,}06\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17\text{ g/mol}}\approx117647{,}0\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17\text{ g/mol}}\approx117647{,}\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17\text{ g/mol}}\approx117647\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17\text{ g/mol}}\approx11764\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17\text{ g/mol}}\approx1176\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17\text{ g/mol}}\approx117\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17\text{ g/mol}}\approx1174\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17\text{ g/mol}}\approx11743\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17\text{ g/mol}}\approx117439\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17\text{ g/mol}}\approx117439{,}\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17\text{ g/mol}}\approx117439{,}8\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17\text{ g/mol}}\approx117439{,}81\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17{,}\text{ g/mol}}\approx117439{,}81\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17{,}0\text{ g/mol}}\approx117439{,}81\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17{,}03\text{ g/mol}}\approx117439{,}81\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17{,}03\text{ g/mol}}\approx117439{,}8\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17{,}03\text{ g/mol}}\approx117439{,}\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17{,}03\text{ g/mol}}\approx117439\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17{,}03\text{ g/mol}}\approx11743\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17{,}03\text{ g/mol}}\approx1174\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17{,}03\text{ g/mol}}\approx117\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17{,}03\text{ g/mol}}\approx1176\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17{,}03\text{ g/mol}}\approx11764\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17{,}03\text{ g/mol}}\approx117647\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17{,}0\text{ g/mol}}\approx117647\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000\text{ g}}{17{,}\text{ g/mol}}\approx117647\text{ mol}n=\frac{massa}{molaire massa}=n_{NH3}=\frac{2.000.000 \text{ g}}{17 \text{ g/mol}}\approx117647\text{ mol}n=\frac{massa}{molaire massa}=_{NH3}=\frac{2.000.000 \text{ g}}{17 \text{ g/mol}}\approx117647\text{ mol}n=\frac{massa}{molaire massa}=N_{NH3}=\frac{2.000.000 \text{ g}}{17 \text{ g/mol}}\approx117647\text{ mol}=\frac{massa}{molaire massa}=N_{NH3}=\frac{2.000.000 \text{ g}}{17 \text{ g/mol}}\approx117647\text{ mol}N=\frac{massa}{molaire massa}=N_{NH3}=\frac{2.000.000 \text{ g}}{17 \text{ g/mol}}\approx117647\text{ mol}N=\frac{massa}{molaire massa}=N_{NH3}=\frac{2.000.000 \text{ g}}{17 \text{ g/mol}}\approx11764\text{ mol}N=\frac{massa}{molaire massa}=N_{NH3}=\frac{2.000.000 \text{ g}}{17 \text{ g/mol}}\approx1176\text{ mol}N=\frac{massa}{molaire massa}=N_{NH3}=\frac{2.000.000 \text{ g}}{17 \text{ g/mol}}\approx117\text{ mol}N=\frac{massa}{molaire massa}=N_{NH3}=\frac{2.000.000 \text{ g}}{17 \text{ g/mol}}\approx1173\text{ mol}N=\frac{massa}{molaire massa}=N_{NH3}=\frac{2.000.000 \text{ g}}{17 \text{ g/mol}}\approx11737\text{ mol}N=\frac{massa}{molaire massa}=N_{NH3}=\frac{2.000.000 \text{ g}}{17 \text{ g/mol}}\approx117371\text{ mol}N=\frac{massa}{molaire massa}N_{NH3}=\frac{2.000.000 \text{ g}}{17 \text{ g/mol}}\approx117371\text{ mol}N=\frac{massa}{molaire massa}[N_{NH3}=\frac{2.000.000 \text{ g}}{17 \text{ g/mol}}\approx117371\text{ mol}

Stap B: Bereken het aantal mol waterstof

De molecuulverhouding in de reactievergelijking laat zien dat 2 mol ammoniak overeenkomt met 3 mol waterstof:

1.Stel de verhouding op: \frac{3}{2}=\frac{x}{117439{,}81}\frac{3}{2}=\frac{x}{}\frac{3}{2}=\frac{x}{1}\frac{3}{2}=\frac{x}{11}\frac{3}{2}=\frac{x}{117}\frac{3}{2}=\frac{x}{1176}\frac{3}{2}=\frac{x}{11764}\frac{3}{2}=\frac{x}{117647}\frac{3}{2}=\frac{x}{117647{,}}\frac{3}{2}=\frac{x}{117647{,}0}\frac{3}{2}=\frac{x}{117647{,}06}\frac{3}{2}=\frac{x}{117647{,}0}\frac{3}{2}=\frac{x}{117647{,}}\frac{3}{2}=\frac{x}{117647}\frac{3}{2}=\frac{x}{11764}\frac{3}{2}=\frac{x}{1176}\frac{3}{2}=\frac{x}{117}\frac{3}{2}=\frac{x}{1174}\frac{3}{2}=\frac{x}{11743}\frac{3}{2}=\frac{x}{117439}\frac{3}{2}=\frac{x}{117439{,}}\frac{3}{2}=\frac{x}{117439{,}8}\frac{3}{2}=\frac{x}{117439{,}81}\frac{3}{2}=\frac{x}{}\frac{3}{2}=\frac{x}{1}\frac{3}{2}=\frac{x}{11}\frac{3}{2}=\frac{x}{117}\frac{3}{2}=\frac{x}{1176}\frac{3}{2}=\frac{x}{11764}\frac{3}{2}=\frac{x}{117647}\frac{3}{2}=\frac{x}{11764}\frac{3}{2}=\frac{x}{1176}\frac{3}{2}=\frac{x}{117}\frac{3}{2}=\frac{x}{1173}\frac{3}{2}=\frac{x}{11737} Hierin is x het aantal mol waterstof.

2.Los de vergelijking op:

x=\frac32{}{}\times117439{,}81\approx176159{,}71\text{ mol H}_2x=\frac32{}{}117439{,}81\approx176159{,}71\text{ mol H}_2x=\frac32{}{}\cdot117439{,}81\approx176159{,}71\text{ mol H}_2x=\frac32{}{}117439{,}81\approx176159{,}71\text{ mol H}_2x=\frac32{}{}\times117439{,}81\approx176159{,}71\text{ mol H}_2x=\frac{3}{\placeholder{}}{}{}\times117439{,}81\approx176159{,}71\text{ mol H}_2x=\frac{}{\placeholder{}}{}{}\times117439{,}81\approx176159{,}71\text{ mol H}_2x=\frac{1}{\placeholder{}}{}{}\times117439{,}81\approx176159{,}71\text{ mol H}_2x={}{}\times117439{,}81\approx176159{,}71\text{ mol H}_2x=r{}{}\times117439{,}81\approx176159{,}71\text{ mol H}_2x=rac{3}{}\times117439{,}81\approx176159{,}71\text{ mol H}_2

Stap C: Bepaal de massa waterstof

1.Bepaal de molaire massa van waterstof (H_2H):

H=1{,}01\;g/molH=1{,}0\;g/molH=1{,}\;g/molH=1\;g/mol, dus H_2=2{,}02\;g/molH_2=2{,}0\;g/molH_2=2{,}\;g/molH_2=2\;g/molH_2=2{,}\;g/molH_2=2{,}0\;g/molH_2=2{,}02\;g/molH_2=2{,}0\;g/molH_2=2{,}\;g/molH_2=2\;g/molH=2\;g/molH2=2\;g/mol.

2. Bereken de totale massa van de benodigde waterstof: massa=aantal\;mol\times molairemassamassa=aantal\;mo\times molairemassamassa=aantal\;m\times molairemassamassa=aantal\;\times molairemassamassa=aantal\times molairemassamassa=aantalm\times molairemassamassa=aantalmo\times molairemassamassa=aantalmol\times molairemassamassa=aantalmo\times molairemassamassa=aantalm\times molairemassamassa=aantal\times molairemassamassa=aanta\times molairemassamassa=aantal\times molairemassamassa=aantal\text{ }\times molairemassamassa=aantal\text{ l}\times molairemassamassa=aantal\text{ ml}\times molairemassa massa\;H_2=176159{,}71\;mol\times2{,}02\;g/mol=355842{,}61\;gmassa\;H_2=176159{,}71\;mol\times2{,}02\;g/mol=355842{,}6\;gmassa\;H_2=176159{,}71\;mol\times2{,}02\;g/mol=355842{,}\;gmassa\;H_2=176159{,}71\;mol\times2{,}02\;g/mol=355842\;gmassa\;H_2=176159{,}71\;mol\times2{,}02\;g/mol=35584\;gmassa\;H_2=176159{,}71\;mol\times2{,}02\;g/mol=3558\;gmassa\;H_2=176159{,}71\;mol\times2{,}02\;g/mol=355\;gmassa\;H_2=176159{,}71\;mol\times2{,}02\;g/mol=35\;gmassa\;H_2=176159{,}71\;mol\times2{,}02\;g/mol=3\;gmassa\;H_2=176159{,}71\;mol\times2{,}02\;g/mol=35\;gmassa\;H_2=176159{,}71\;mol\times2{,}02\;g/mol=356\;gmassa\;H_2=176159{,}71\;mol\times2{,}02\;g/mol=3564\;gmassa\;H_2=176159{,}71\;mol\times2{,}02\;g/mol=35647\;gmassa\;H_2=176159{,}71\;mol\times2{,}02\;g/mol=356470\;gmassa\;H_2=176159{,}71\;mol\times2{,}02\;g/mol=356470{,}\;gmassa\;H_2=176159{,}71\;mol\times2{,}02\;g/mol=356470{,}5\;gmassa\;H_2=176159{,}71\;mol\times2{,}02\;g/mol=356470{,}59\;gmassa\;H_2=\;mol\times2{,}02\;g/mol=356470{,}59\;gmassa\;H_2=1\;mol\times2{,}02\;g/mol=356470{,}59\;gmassa\;H_2=17\;mol\times2{,}02\;g/mol=356470{,}59\;gmassa\;H_2=176\;mol\times2{,}02\;g/mol=356470{,}59\;gmassa\;H_2=1764\;mol\times2{,}02\;g/mol=356470{,}59\;gmassa\;H_2=17647\;mol\times2{,}02\;g/mol=356470{,}59\;gmassa\;H_2=176470\;mol\times2{,}02\;g/mol=356470{,}59\;gmassa\;H_2=176470{,}\;mol\times2{,}02\;g/mol=356470{,}59\;gmassa\;H_2=176470{,}5\;mol\times2{,}02\;g/mol=356470{,}59\;gmassa\;H_2=176470{,}59\;mol\times2{,}02\;g/mol=356470{,}59\;gmassa\;H_2=176470{,}59\;mol\times2{,}02\;g/mol=356470{,}5\;gmassa\;H_2=176470{,}59\;mol\times2{,}02\;g/mol=356470{,}\;gmassa\;H_2=176470{,}59\;mol\times2{,}02\;g/mol=356470\;gmassa\;H_2=176470{,}59\;mol\times2{,}02\;g/mol=35647\;gmassa\;H_2=176470{,}59\;mol\times2{,}02\;g/mol=3564\;gmassa\;H_2=176470{,}59\;mol\times2{,}02\;g/mol=356\;gmassa\;H_2=176470{,}59\;mol\times2{,}02\;g/mol=35\;gmassa\;H_2=176470{,}59\;mol\times2{,}02\;g/mol=355\;gmassa\;H_2=176470{,}59\;mol\times2{,}02\;g/mol=3558\;gmassa\;H_2=176470{,}59\;mol\times2{,}02\;g/mol=35584\;gmassa\;H_2=176470{,}59\;mol\times2{,}02\;g/mol=355842\;gmassa\;H_2=176470{,}59\;mol\times2{,}02\;g/mol=355842{,}\;gmassa\;H_2=176470{,}59\;mol\times2{,}02\;g/mol=355842{,}6\;gmassa\;H_2=176470{,}59\;mol\times2{,}02\;g/mol=355842{,}63\;gmassa\;H_2=\;mol\times2{,}02\;g/mol=355842{,}63\;gmassa\;H_2=1\;mol\times2{,}02\;g/mol=355842{,}63\;gmassa\;H_2=17\;mol\times2{,}02\;g/mol=355842{,}63\;gmassa\;H_2=176\;mol\times2{,}02\;g/mol=355842{,}63\;gmassa\;H_2=1761\;mol\times2{,}02\;g/mol=355842{,}63\;gmassa\;H_2=17615\;mol\times2{,}02\;g/mol=355842{,}63\;gmassa\;H_2=176159\;mol\times2{,}02\;g/mol=355842{,}63\;gmassa\;H_2=176159{,}\;mol\times2{,}02\;g/mol=355842{,}63\;gmassa\;H_2=176159{,}7\;mol\times2{,}02\;g/mol=355842{,}63\;gmassa\;H_2=176159{,}72\;mol\times2{,}02\;g/mol=355842{,}63\;gmassa\;H_2=176159{,}72\;mol\times2{,}02\;g/mol=355842{,}6\;gmassa\;H_2=176159{,}72\;mol\times2{,}02\;g/mol=355842{,}\;gmassa\;H_2=176159{,}72\;mol\times2{,}02\;g/mol=355842\;gmassa\;H_2=176159{,}72\;mol\times2{,}02\;g/mol=35584\;gmassa\;H_2=176159{,}72\;mol\times2{,}02\;g/mol=3558\;gmassa\;H_2=176159{,}72\;mol\times2{,}02\;g/mol=355\;gmassa\;H_2=176159{,}72\;mol\times2{,}02\;g/mol=35\;gmassa\;H_2=176159{,}72\;mol\times2{,}02\;g/mol=352\;gmassa\;H_2=176159{,}72\;mol\times2{,}02\;g/mol=3529\;gmassa\;H_2=176159{,}72\;mol\times2{,}02\;g/mol=35294\;gmassa\;H_2=176159{,}72\;mol\times2{,}02\;g/mol=352940\;gmassa\;H_2=176159{,}72\;mol\times2{,}0\;g/mol=352940\;gmassa\;H_2=176159{,}72\;mol\times2{,}\;g/mol=352940\;gmassa\;H_2=176159{,}72\;mol\times2\;g/mol=352940\;gmassa\;H_2=176470{,}5\;mol\times2\;g/mol=352940\;gmassa\;H_2=176470{,}5\;mol\times2\;g/mol=35294\;gmassa\;H_2=176470{,}5\;mol\times2\;g/mol=3529\;gmassa\;H_2=176470{,}5\;mol\times2\;g/mol=35293\;gmassa\;H_2=176470{,}5\;mol\times2\;g/mol=3529\;gmassa\;H_2=176470{,}5\;mol\times2\;g/mol=352\;gmassa\;H_2=176470{,}5\;mol\times2\;g/mol=35\;gmassa\;H_2=176470{,}5\;mol\times2\;g/mol=356\;gmassa\;H_2=176470{,}5\;mol\times2\;g/mol=3564\;gmassa\;H_2=176470{,}5\;mol\times2\;g/mol=35647\;gmassa\;H_2=176470{,}5\;mol\times2\;g/mol=356470\;gmassa\;H_2=176470{,}5\;mol\times2\;g/mol=356470{,}\;gmassa\;H_2=176470{,}5\;mol\times2\;g/mol=356470{,}4\;gmassa\;H_2=176470{,}5\;mol\times2\;g/mol=356470{,}41\;gmassa\;H_2=176470{,}5\;mol\times2{,}\;g/mol=356470{,}41\;gmassa\;H_2=176470{,}5\;mol\times2{,}0\;g/mol=356470{,}41\;gmassa\;H_2=176470{,}5\;mol\times2{,}02\;g/mol=356470{,}41\;gmassa\;H_2=176470{,}5\;mol\times2{,}02\;g/mol=356470{,}4\;gmassa\;H_2=176470{,}5\;mol\times2{,}02\;g/mol=356470{,}\;gmassa\;H_2=176470{,}5\;mol\times2{,}02\;g/mol=356470\;gmassa\;H_2=176470{,}5\;mol\times2{,}02\;g/mol=3564704\;gmassa\;H_2=176470{,}5\;mol\times2{,}02\;g/mol=35647041\;gmassa\;H_2=176470{,}5\;mol\times2{,}02\;g/mol=3564704\;gmassa\;H_2=176470{,}5\;mol\times2{,}02\;g/mol=356470\;gmassa\;H_2=176470{,}5\;mol\times2{,}02\;g/mol=35647\;gmassa\;H_2=176470{,}5\;mol\times2{,}02\;g/mol=3564\;gmassa\;H_2=176470{,}5\;mol\times2{,}02\;g/mol=356\;gmassa\;H_2=176470{,}5\;mol\times2{,}02\;g/mol=3563\;gmassa\;H_2=176470{,}5\;mol\times2{,}02\;g/mol=35633\;gmassa\;H_2=176470{,}5\;mol\times2{,}02\;g/mol=35633{,}\;gmassa\;H_2=176470{,}5\;mol\times2{,}02\;g/mol=35633{,}1\;gmassa\;H_2=176470{,}5\;mol\times2{,}02\;g/mol=35633{,}12\;gmassa\;H_2=176470{,}\;mol\times2{,}02\;g/mol=35633{,}12\;gmassa\;H_2=176470\;mol\times2{,}02\;g/mol=35633{,}12\;gmassa\;H_2=17647\;mol\times2{,}02\;g/mol=35633{,}12\;gmassa\;H_2=1764\;mol\times2{,}02\;g/mol=35633{,}12\;gmassa\;H_2=176\;mol\times2{,}02\;g/mol=35633{,}12\;gmassa\;H_2=17\;mol\times2{,}02\;g/mol=35633{,}12\;gmassa\;H_2=176\;mol\times2{,}02\;g/mol=35633{,}12\;gmassa\;H_2=1760\;mol\times2{,}02\;g/mol=35633{,}12\;gmassa\;H_2=17605\;mol\times2{,}02\;g/mol=35633{,}12\;gmassa\;H_2=176056\;mol\times2{,}02\;g/mol=35633{,}12\;gmassa\;H_2=176056\;mol\times2{,}02\;g/mol=35633{,}1\;gmassa\;H_2=176056\;mol\times2{,}02\;g/mol=35633{,}\;gmassa\;H_2=176056\;mol\times2{,}02\;g/mol=35633\;gmassa\;H_2=176056\;mol\times2{,}02\;g/mol=3563\;gmassa\;H_2=176056\;mol\times2{,}02\;g/mol=356\;gmassa\;H_2=176056\;mol\times2{,}02\;g/mol=3561\;gmassa\;H_2=176056\;mol\times2{,}02\;g/mol=35611\;gmassa\;H_2=176056\;mol\times2{,}02\;g/mol=356112\;gmassa\;H_2=176056\;mol\times2{,}02\;g/mol=35112\;gmassa\;H_2=176056\;mol\times2{,}02\;g/mol=352112\;gmassa\;H_2=176056\;mol\times2{,}0\;g/mol=352112\;gmassa\;H_2=176056\;mol\times2{,}01\;g/mol=352112\;gmassa\;H_2=176056\;mol\times2{,}0\;g/mol=352112\;gmassa\;H_2=176056\;mol\times2{,}\;g/mol=352112\;gmassa\;H_2=176056\;mol\times2\;g/mol=352112\;gmassa\;H_2=176056\;mol\times2\;g/mol=352112gmassa\;H_2=176056\;mol\times2\;g/mol=352112massa\;H_2=176056\;mol\times2\;g/mol=352112gmassa\;H_2=176056\;mol\times2\;g/mol=352112massa\;H_2=176056\;mol\times2\;g/mol=352112\text{ }massa\;H_2=176056\;mol\times2\;g/mol=352112\text{ }massa\;H_2=176056\;mol\times2\;g/mol=352112\text{ g}massa\;H_2=176056\;mol\times2\;g/mol=352112\text{ }massa\;H_2=176056\;mol\times2\;g/mol=352112\text{ }massa\;H_2=176056\;mol\times2\;g/mol=352112\text{ g}massa\;H=176056\;mol\times2\;g/mol=352112\text{ g}massa\;H2=176056\;mol\times2\;g/mol=352112\text{ g}massa\;H2=176056\;mol\times2\;g/mo=352112\text{ g}massa\;H2=176056\;mol\times2\;g/m=352112\text{ g}massa\;H2=176056\;mol\times2\;g/=352112\text{ g}massa\;H2=176056\;mol\times2\;g=352112\text{ g}massa\;H2=176056\;mol\times2\;\frac{g}{\placeholder{}}=352112\text{ g}massa\;H2=176056\;mol\times2\;g=352112\text{ g}massa\;H2=176056\;mol\times2\;g\text{l}=352112\text{ g}massa\;H2=176056\;mol\times2\;g\text{ol}=352112\text{ g}massa\;H2=176056\;mol\times2\;g\text{mol}=352112\text{ g}massa\;H2=176056\;mol\times2\;g\text{/mol}=352112\text{ g}massa\;H2=176056\;mol\times2\;\text{/mol}=352112\text{ g}massa\;H2=176056\;mol\times2\text{/mol}=352112\text{ g}massa\;H2=176056\;mol\times2\text{ /mol}=352112\text{ g}massa\;H2=176056\;mol\times2\text{ g/mol}=352112\text{ g}massa\;H2=176056\;mo\times2\text{ g/mol}=352112\text{ g}massa\;H2=176056\;m\times2\text{ g/mol}=352112\text{ g}massa\;H2=176056\;\times2\text{ g/mol}=352112\text{ g}massa\;H2=176056\times2\text{ g/mol}=352112\text{ g}massa\;H2=176056m\times2\text{ g/mol}=352112\text{ g}massa\;H2=176056\times2\text{ g/mol}=352112\text{ g}massa\;H2=176056\text{ }\times2\text{ g/mol}=352112\text{ g}massa\;H2=176056\text{ m}\times2\text{ g/mol}=352112\text{ g}massa\;H2=176056\text{ mo}\times2\text{ g/mol}=352112\text{ g}massa\;H2=176056\text{ mol}\times2\text{ g/mol}=352112\text{ g} Dat is ongeveer 355{,}84\;kg355{,}8\;kg355{,}85\;kg355{,}8\;kg355{,}\;kg355\;kg35\;kg356\;kg35\;kg353\;kg35\;kg356\;kg35\;kg352\;kg352kg352/kg.