Answer:
C
Explanation:
We want to calculate the volume of carbon dioxide gas produced at STP upon the decomposition of 0.150 g of CaCO₃ illustrated by the equation:
[tex]\displaystyle \text{CaCO$_3$(s)} \longrightarrow \text{CaO(s)} + \text{CO$_2$(g)}[/tex]
To do so, we can convert from grams of CaCO₃ to moles of CaCO₃; moles of CaCO₃ to moles of CO₂; and moles of CO₂ to liters of CO₂ (at STP).
Find the molecular weight of CaCO₃:
[tex]\displaystyle \begin{aligned}\mathcal{M}_\text{CaCO$_3$} & = (40.08 + 12.01 + 3(16.00)) \text{ g/mol} \\ \\ & = 100.09 \text{ g/mol} \end{aligned}[/tex]
From the equation, one mole of CO₂ is produced from every one mole of CaCO₃.
Finally, recall that at STP, one mole of any gas occupies a volume of 22.4 L.
Hence:
[tex]\displaystyle \begin{aligned} 0.150 \text{ g CaCO$_3$} & \cdot \frac{1 \text{ mol CaCO$_3$}}{100.09\text{ g CaCO$_3$}} \cdot \frac{1 \text{ mol CO$_2$}}{1 \text{ mol CaCO$_3$}}\cdot \frac{22.4 \text{ L CO$_2$}}{1\text{ mol CO$_2$}} \\ \\ &= 0.0336 \text{ L CO$_2$}\end{aligned}[/tex]
In conclusion, about 0.0336 liters of carbon dioxide is produced.
Our answer is C.
