Calulate the mass of copper required to produce 159 g of copper (ll) oxide when heated in excess oxygen.

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Solution

$2 C u + O_{2} ⟶ C u O$
Given mass of CuO= $159 g$
$1$ mole of Cu $⟶ 63.5 g$
$1$ mole of CuO $⟶ 79.5 g$
From the reaction
$2$ mol of Cu reacts with $2$ moles of CuO
$1$ mol of Cu $⟶ 79.5 g$
Thus $2$ mol of CuO $⟶ 159 g$
$159 g$ of CuO $⟶ 2$ mol of Cu
$2$ mol of Cu $⟶ 2$ x $63.5 g$ of Cu
$⟶ 127 g$ of Cu
Therefore, the mass of Cu required to produce $159 g$ of CuO when heated in excess oxygen is $127 g$

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