# A sample of dolomitic limestone containing only CaCO_3 and MgCO_3 was analyzed. When a 0.2800 gram sample of this limestone was decomposed by heating, 75.0 mL of CO_2 at 750 mmHG and 20°C were evolved. How many grams of CO_2 were produced?

Jan 8, 2017

$M C {O}_{3} \left(s\right) + \Delta \rightarrow M O \left(s\right) + C {O}_{2} \left(g\right) \uparrow$

Approx. $140 \cdot m g$ gas are evolved.

#### Explanation:

We use the Ideal Gas equation to access the moles of carbon dioxide evolved:

$n = \frac{P V}{R T}$ $=$ $\frac{\frac{750 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t {m}^{-} 1} \times 75.0 \times {10}^{-} 3 \cdot L}{0.0821 \cdot L \cdot a t m \cdot {K}^{-} 1 \cdot m o {l}^{-} 1 \times 293 \cdot K}$

$= 3.1 \times {10}^{-} 3 m o l$.

"Mass of carbon dioxide produced"=3.1xx10^-3molxx44.0*g*mol^-1=??g

The key to doing this problem was (i) knowing that $760 \cdot m m \cdot H g \equiv 1 \cdot a t m$; and (ii) using the right gas constant, $R$. I have always liked $R = 0.0821 \cdot L \cdot a t m \cdot {K}^{-} 1 \cdot m o {l}^{-} 1$ because it uses units that chemists would habitually use. The other gas constants tend to use whack, non-intuitive units.