# Question #f8d64

Oct 25, 2014

You must first be sure the equation is balanced, which it is. Then you must determine the limiting reactant, as that will determine the theoretical yield of calcium carbonate.

$\text{CaO} \left(s\right)$ + ${\text{CO}}_{2} \left(g\right)$ $\rightarrow$ ${\text{CaCO}}_{3} \left(s\right)$

The mole ratios for this equation are 1:1.

Determine the number of moles of each reactant. Moles are calculated by dividing the mass of each reactant by its molar mass. The molar mass of $\text{CaO(s)" = "56.077 g/mol}$. The molar mass of
$\text{CO"_2(g)}$ = $\text{44.099g/mol}$

Moles of CaO.

$\text{14.4g CaO(s)}$ x $\text{1 mol CaO(s)"/"56.077 g/mol}$ = $\text{0.257 mol CaO(s)}$

Moles of $\text{CO"_2(g)}$.

${\text{13.8g CO}}_{2} \left(g\right)$ x $\text{1 mol"/"44.099g/mol}$ = ${\text{0.313 mol CO}}_{2}$

Since there is less $\text{CaO(s)}$, it is the limiting reactant.

Since the mole ratio of $\text{CaO(s)}$ and the product ${\text{CaCO}}_{3} \left(s\right)$ is 1:1, the reaction can produce no more than ${\text{0.257 mol CaCO}}_{3} \left(s\right)$

To determine the theoretical yield of ${\text{CaCO}}_{3} \left(s\right)$ in grams, multiply the number of moles of ${\text{CaCO}}_{3} \left(s\right)$ times its molar mass.

${\text{0.257 mol CaCO}}_{3} \left(s\right)$ x $\text{100.11g"/"1 mol}$ = ${\text{25.72g CaCO}}_{3} \left(s\right)$

The theoretical yield of ${\text{CaCO}}_{3} \left(s\right)$ in this experiment is $\text{25.72g}$