Question

The reaction of iron ore with carbon follows the equation: 2Fe2O3 + 3C + 3C02 → 4Fe In a reaction, 500 g of Fe2O3 are mixed with 100 g of C. The actual yield of Fe is 5.0 g, what is the percent yield of iron? o 40% 0 60% 20% O 80 % 10 %

Answer 1

Balancing equation is,

2Fe2CO3 + 3C = 4Fe + 3CO2

500 g of Fe2CO3 used and its molecular weight = 159.7 g/mol, hence its moles = wt / mol wt

500 g / 159.7 g/mol = 3.1308 moles Fe2CO3 used in reation

100 g of C used and its molecular weight = 12.01 g/mol, hence its moles = wt / mol wt

100 g / 12.01 g/mol = 8.3263 moles C used in reation.

From balanced equation it is clear that 2 moles of Fe2CO3 required 3 moles of Carbon.

Then 3.1308 moles Fe2CO3  required = 3.1308 * 3 / 2

= 9.3924 / 2

= 4.6962 moles of C required. Hence carbon (C) is uesd in excess and limiting reagent is  Fe2CO3

From balanced equation,

2 moles of Fe2CO3 will produced 4 moles of Fe.

Then  3.1308 moles Fe2CO3 are produced = 3.1308 * 4 / 2

- 12.5232 / 2

= 6.2616 moles of iron will produced as 100% yield considered.

Mol wt of Fe = 55.85 g/mol

Theoretical wt of Fe = 6.2616 mol * 55.85 g/mol

= 349.7 g of Fe 100% yield (Theoretical yield)

Parcent yield = Actual yield / Theoretical yield * 100

= 5 / 349.7 * 100

= 1.43%

Hence Actual parcent yield of Iron (Fe) = 1.43%