Question

Part 1:

Iron can be produced by magnetite according to the following reaction:

___Fe3O4 + ___H2 → ___Fe + ___H2O

Balance the reaction, sum up the reaction (including implied ones), and enter the sum below.

Part 2:

33.08 grams of Fe3O4 are combined with 1.43 grams of H2. If all of the Fe3O4reacts, how many moles of Fe could be formed?

Part 3:

If, instead, all of the hydrogen (H2) reacts, how many moles of Fe could be formed?

Part 4:

Determine which substance (Fe3O4 or H2) is the limiting reactant. After, enter in the space below how many grams of Iron (Fe) could theoretically be formed?

Part 5:

If 16.844 grams of iron (Fe) are obtained, what was the percent yield of the reaction?

Answer 1

1) Fe3O4 + 4H2 <---------> 3Fe + 4H2O

2) 231.53 g Fe2O3 forms 3 mole Fe

33.08 g Fe2O3 forms 33.08 x 3 / 231.53 = 0.43 moles Fe

moles of Fe formed = 0.43 mole

3) 4 x 2 g H2 forms 3 moles fe

1.43 g H2 forms 1.43 x 3 / 4 x 2 = 0.54 moles Fe

moles of Fe formed = 0.54 moles

4) 231.53 g Fe2O3 reacts with 4 x 2 g H2

33.08 g fe2O3 reacts with 33.08 x 4 x 2 / 231.53 = 1.14 g H2

but we have 1.43 g H2. so H2 is exess reagent

Fe2O3 is limiting reagent.

5) convert moles of Fe to mass

mass of Fe = 0.43 x 55.84 = 24 g

% yield = (16.844 / 24) x 100

% yield = 70.18