Question

Calculate the volume of carbon dioxide gas in liters that is produced at STP upon the decomposition of 0.150 g CaCO3 via the equation: CaCO3(s) → Cao (s) + CO2(g) 3 O

a. 1.31
b. 0.0196 L
c. 0.0336 L
d. 2.24 L
e. 6.31 L
​

Answer 1

C

Step-by-step 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:

`\displaystyle \text{CaCO$_3$(s)} \longrightarrow \text{CaO(s)} + \text{CO$_2$(g)}`

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:

`\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}`

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:

`\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}`

In conclusion, about 0.0336 liters of carbon dioxide is produced.

Our answer is C.