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Fofino [41]
3 years ago
10

2Fe(s) +3H2SO4(aq) →Fe2(SO4)3(aq) +3H2(g)When 10.3 g of iron are reacted with 14.8 moles of sulfuric acid, what is the percent y

ield if 5.40 g of "hydrogen gas" are collected?
Chemistry
1 answer:
Elden [556K]3 years ago
7 0

Answer:

1040%

Explanation:

To solve this question we must convert the mass of Iron to moles in order to find limiting reactant. With limiting reactant we can find the theoretical moles of hydrogen and theoretical mass:

Percent yield = Actual yield (5.40g) / Theoretical yield * 100

<em>Moles Fe -Molar mass: 55.845g/mol-:</em>

10.3g * (1mol / 55.845g) = 0.184 moles of Fe will react.

For a complete reaction of these moles there are necessaries:

0.184 moles Fe* ( 3 mol H2SO4 / 2 mol Fe) = 0.277 moles H2SO4.

As there are 14.8 moles of the acid, <em>Fe is limiting reasctant.</em>

The moles of H2 produced are:

0.184 moles Fe* ( 3 mol H2 / 2 mol Fe) = 0.277 moles H2

The mass is:

0.277 moles H2 * (2.016g/mol) = 0.558g H2

Percent yield is:

5.40g / 0.558g * 100 = 1040%

It is possible the experiment wasn't performed correctly

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