# Question #3f1d9

Apr 4, 2014

For this solution, since the reaction is not at Standard Temperature and Pressure (STP), you will need to use the Ideal Gas Law to solve for moles of $C {O}_{2}$ and then Stoichiometry to solve for the grams of $C a C {O}_{3}$.

To begin with we find the knowns and unknowns for the Ideal Gas Law, PV = nRT.

P = 1.00 atm
V = 3.45 L
n = ??
R = 0.0821 $\frac{a t m L}{m o l K}$
T = 298 K

PV = nRT can be rearranged $n = \frac{P V}{R T}$

$n = \frac{\left(1.00 a t m\right) \left(3.45 L\right)}{\left(.0821 \frac{a t m L}{m o l K}\right) \left(298 K\right)}$

n = 0.14 moles $C {O}_{2}$

Now you can use the balanced chemical equation and stoichiometry to solve for the grams of $C a C {O}_{3}$

moles $\to$ moles $\to$ grams

$0.14 m o l C {O}_{2} x \frac{1 m o l C a C {O}_{3}}{1 m o l C {O}_{2}} x \frac{100.0869 g C a C {O}_{3}}{1 m o l C a C {O}_{3}} = 14.01 g C a C {O}_{3}$

I hope this was helpful.
SMARTERTEACHER