Calcium carbonate decomposes to form calcium oxide and carbon dioxide gas. If 44.5 grams of calcium oxide is produced, how many grams of calcium carbonate had to decomposez?

Mar 1, 2017

$\text{Approx.}$ $80 \cdot g$

Explanation:

We need a stoichiometric reaction that represents the decomposition of calcium carbonate:

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

And thus calcium oxide and calcium carbonate are present in equimolar amounts:

$\text{Moles of calcium oxide}$ $=$ $\frac{44.5 \cdot g}{56.08 \cdot g \cdot m o {l}^{-} 1} = 0.794 \cdot m o l$.

Given the stoichiometry of the reaction, an equimolar quantity of calcium carbonate MUST have been present.

This represents a mass of $0.794 \cdot m o l \times 100.09 \cdot g \cdot m o {l}^{-} 1$

$\cong 80 \cdot g$

In the decomposition reaction, what does the $\Delta$ symbol represent?