2Fe(s) +3H2SO4(aq) Fe2(SO4)3(aq) +3H2(g) How many moles of hydrogen gas can be p

Question

2Fe(s) +3H2SO4(aq) Fe2(SO4)3(aq) +3H2(g)

How many moles of hydrogen gas can be produced from 30.0 g of iron?

When 10.3 g of iron are reacted with 14.8 moles of sulfuric acid, what is what is the limiting reactant?

When 10.3 g of iron are reacted with 14.8 moles of sulfuric acid, what is the maximum mass of hydrogen gas that can be produced?

)When 10.3 g of iron are reacted with 14.8 moles of sulfuric acid, what is the percent yield if 5.40 g of "hydrogen gas" are collected?

When 10.3 g of iron are reacted with 14.8 moles of sulfuric acid, if 5.40 g of "hydrogen gas" are collected is it likely that the experiment was performed correctly?

Explanation / Answer

2Fe(s) +3H2SO4(aq) Fe2(SO4)3(aq) +3H2(g)

3 moles H2 is produced from 2 moles of Fe

30g Fe = 30g/56g/mol = 0.54 moles

Moles of H2 formed = (3/2) *0.54 = 0.80 moles

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Moles of Fe used = 10.3gm/56g/mol = 10.3g/56g/mol = 0.184 mol

Moles of H2SO4 used = 14.8 mol

According to the balanced reaction above , 0.184 mol Fe reacts with (3/2) * 0.184 mol = 0.276 mol H2SO4. But 14.8 mol H2SO4 is present. So, Fe will be completely consumed and this is the limiting reagent.

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moles of H2 formed = (3/2) * 0.184 mol = 0.276 mol

Mass of H2 formed = 2 g/mol* 0.15 mol = 0.552 gm

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% yield = (I actual- theoretical/theoretical I) *100% = 484.8 %

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Experiment is not performed correctly. More than 100% yield is not possible

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