# Question #e5006

Dec 25, 2017

$= 96.2 g C a C {O}_{3}$

#### Explanation:

1. Write the skeleton and balance the given equation
$C a {\left(O H\right)}_{2} \left(s\right) + C {O}_{2} \left(g\right) > C a C {O}_{3} \left(s\right) + {H}_{2} O \left(g\right)$
2. Find the molar mass of $C a C {O}_{3}$ as the involved compound as specified in the problem.
$= \frac{100 g}{m o l}$
3. Since the given data is already expressed in $m o l$, conversion per convention is no longer necessary in this case.
4. Referring to the balanced equation shown above for the mole ratio, find the $\eta C a C {O}_{3}$ through molar conversion; i.e.,
$= 0.962 {\cancel{m o l C a \left(O H\right)}}_{2} \times \frac{1 m o l C a C {O}_{3}}{1 \cancel{m o l C a {\left(O H\right)}_{2}}}$
$= 0.962 m o l C a C {O}_{3}$
5. Now, find the $\text{mass} \left(m\right)$ of $C a C {O}_{3}$, knowing the $\eta C a C {O}_{3}$ as computed, and the molar mass of the compound for the conversion.
$= 0.962 \cancel{m o l C a C {O}_{3}} \times \frac{100 g C a C {O}_{3}}{1 \cancel{m o l C a C {O}_{3}}}$
$= 96.2 g C a C {O}_{3}$