What mass of \(\ce{Cu(s)}\) would be produced if 0.40 mol of \(\ce{Cu2O(s)}\) was reduced completely with excess \(\ce{H2(g)}\) ?

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We first need to write the balanced equation. The oxidation state of copper changes from +1 to 0.

$$ \ce{Cu2O(s) ->[{reduction}] Cu(s) } $$

The other reactant must be oxidized. The oxidation state of hydrogen will increase. Water will be the likely product since we need a compound with hydrogen and oxygen.

$$ \ce{Cu2O(s) + H2(g) -> H2O + Cu(s) } $$

Balancing this equation:

$$ \ce{Cu2O(s) + H2(g) -> H2O + 2Cu(s) } $$

Using stoichiometry:

$$ \pu{0.40 mol } \ce{Cu2O(s)}\cdot \frac{\pu{2 mol } \ce{Cu(s)}}{\pu{1 mol } \ce{Cu2O(s)}} \cdot \frac{\pu{63.5 g }\ce{Cu(s)}}{\pu{1 mol } \ce{Cu(s)}} $$ $$ \approx \boxed{\pu{51 g } \ce{Cu(s)}} $$