Question

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what volume of carbon dioxide is produced when 64.4 g of calcium carbonate reacts completely according to the following reaction at 25°c and 1 atm?
calcium carbonate (s) → calcium oxide (s) + carbon dioxide (g)
volume =
l
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Explanation:

Step1: Calculate the molar mass of calcium carbonate ($CaCO_3$)

The molar mass of $Ca$ is $40.08\ g/mol$, $C$ is $12.01\ g/mol$, and $O$ is $16.00\ g/mol$.
$Molar\ mass\ of\ CaCO_3=40.08 + 12.01+3\times16.00=100.09\ g/mol$

Step2: Calculate the number of moles of calcium carbonate

$n=\frac{m}{M}$, where $m = 64.4\ g$ and $M = 100.09\ g/mol$
$n=\frac{64.4\ g}{100.09\ g/mol}\approx0.643\ mol$

Step3: Determine the mole ratio from the balanced chemical equation

The balanced chemical equation is $CaCO_3(s)
ightarrow CaO(s)+CO_2(g)$. The mole ratio of $CaCO_3$ to $CO_2$ is $1:1$. So, $n(CO_2)=n(CaCO_3) = 0.643\ mol$

Step4: Use the ideal gas law $PV = nRT$ to find the volume of $CO_2$

Given $P = 1\ atm$, $n=0.643\ mol$, $R = 0.0821\ L\cdot atm/(mol\cdot K)$, and $T=(25 + 273.15)K=298.15\ K$
$V=\frac{nRT}{P}$
$V=\frac{0.643\ mol\times0.0821\ L\cdot atm/(mol\cdot K)\times298.15\ K}{1\ atm}$
$V=\frac{0.643\times0.0821\times298.15}{1}\ L$
$V\approx15.7\ L$

Answer:

$15.7$