Question

Calculate the mass of copper required to produce ${\text{159 g}}$ of copper (II) oxide when heated in excess oxygen.

Answer 1

Hint: Copper is heated in excess oxygen to form copper (II) oxide. Write a correct balanced chemical equation for the reaction. The chemical formula for copper (II) oxide is ${\text{CuO}}$. The oxidation state of copper is ${\text{ + 2}}$.

Complete answer: 

Write the balanced chemical equation for the reaction as follows:
Copper reacts with oxygen to form copper (II) oxide. 

The reaction is,
${\text{2Cu + }}{{\text{O}}_2}{\text{ }} \to {\text{ 2CuO}}$
Use the reaction for further stoichiometric calculations.
-Calculate the number of moles of copper (II) oxide in ${\text{159 g}}$ of copper (II) oxide using the equation as follows:
${\text{Number of moles of CuO}} = \dfrac{{{\text{Mass of CuO}}}}{{{\text{Molar mass of CuO}}}}$
Substitute ${\text{159 g}}$ for the mass of copper (II) oxide, ${\text{79}}{\text{.5 g mo}}{{\text{l}}^{ - 1}}$ for the molar mass of copper (II) oxide. Thus,
${\text{Number of moles of CuO}} = \dfrac{{{\text{159 g}}}}{{{\text{79}}{\text{.5 g mo}}{{\text{l}}^{ - 1}}}}$
$\Rightarrow {\text{Number of moles of CuO}} = {\text{2 mol}}$
Thus, the number of moles of copper (II) oxide in ${\text{159 g}}$ of copper (II) oxide are ${\text{2 mol}}$.
Calculate the number of moles of copper required to produce ${\text{2 mol}}$ of copper (II) oxide as follows:
From the reaction stoichiometry,
${\text{2 mol}}$ copper is required to produce ${\text{2 mol}}$ copper (II) oxide.
Calculate the mass of copper required to produce ${\text{159 g}}$ of copper (II) oxide as follows:
${\text{1 mol}}$ copper contains ${\text{63}}{\text{.5 g}}$ of copper (molar mass of copper). Thus,
${\text{Mass of copper}} = 2{\text{ mol}} \times \dfrac{{{\text{63}}{\text{.5 g}}}}{{{\text{1 mol}}}}$
$\Rightarrow {\text{Mass of copper}} = {\text{127 g}}$

Thus, the mass of copper required to produce ${\text{159 g}}$ of copper (II) oxide is ${\text{127 g}}$.

Note: The most important step is to write the correct balanced chemical equation, if the balanced chemical equation is not correct, the reaction stoichiometry changes and the final answer changes.