What volume of oxygen gas,measured at STP,would be required for the preparation of 39.0 g of sodium peroxide from sufficient sodium metal?
A)11.2 L
B)9.6 L
C)7.4 L
D)5.6 L
E)2.2 L

Question

Quizplus · Accepted Answer

The balanced chemical equation for the formation of sodium peroxide from sodium metal and oxygen is: $4Na + O_2 \rightarrow 2Na_2O_2$. The molar mass of sodium peroxide ($Na_2O_2$) is approximately 78 g/mol. Therefore, 39.0 g of $Na_2O_2$ is 0.5 moles. According to the stoichiometry of the reaction, 0.5 moles of $O_2$ are required to produce 1 mole of $Na_2O_2$. At STP, 1 mole of any gas occupies 22.4 L, so 0.5 moles of $O_2$ would occupy $22.4 L \times 0.5 = 11.2 L$.

What Volume of Oxygen Gas,measured at STP,would Be Required for the Preparation