Question

Fe2O3 + 3 H2 –> 2 Fe + 3 H2O 6.3 g of Fe2O3 is reacted with 15.8 g of hydrogen to form iron and water. How many moles of the excess reagent are left over?

Answer 1

Step 1

The reaction must be written, completed, and balanced.

Fe2O3 + 3 H2 => 2 Fe + 3 H2O

The reactants: Fe2O3 and H2

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Step 2

Information provided:

The mass of Fe2O3 = 6.3 g

The mass of H2 = 15.8 g

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Information needed:

The molar masses:

Fe2O3) 160 g/mol approx.

H2) 2.01 g/mol approx.

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Step 3

The limiting reactant and the excess agent:

By stoichiometry,

1 mole Fe2O3 = 160 g

1 mole H2 = 2.01 g

Fe2O3 + 3 H2 => 2 Fe + 3 H2O

160 g Fe2O3 ------- 3 x 2.01 g H2

6.3 g Fe2O3 -------- X

X = 6.3 g Fe2O3 x 3 x 2.01 g H2/160 g Fe2O3 = 0.24 g

For 6.3 g Fe2O3, 0.24 g of H2 is needed, but there is 15.8 g. Therefore, the limiting reactant is Fe and the excess is the H2.

Amount of the H2 leftover:

15.8 g (provided) - 0.24 g (needed) = 15.56 g (leftover)

Mass to moles:

15.56 g x (1 mole H2/2.01 g H2) = 7.74 moles

Answer: 7.74 moles of H2 are leftover