# A mass of 2.40*g of calcium oxide is obtained by heating a mass of 5.00*g calcium carbonate. What is the stoichiometric equation that represents the formation of calcium oxide, and what is the percentage yield?

Mar 12, 2017

We need a stoichiometrically balanced equation to represent the decomposition of calcium carbonate..........

#### Explanation:

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

The $\Delta$ symbol represents heat, and you have to fiercely heat these carbonates to effect decomposition. Note that this reaction is certainly stoichiometrically balanced: garbage in equals garbage out.

$\text{Moles of calcium carbonate "=" } \frac{5.00 \cdot g}{100.09 \cdot g \cdot m o {l}^{-} 1}$

$=$ $0.0500 \cdot m o l$

$\text{Moles of calcium oxide "=" } \frac{2.40 \cdot g}{56.08 \cdot g \cdot m o {l}^{-} 1}$

$=$ $0.0428 \cdot m o l$

$\text{% yield}$ $=$ (0.0428*mol)/(0.0500*mol)xx100%=86%

Are you happy with this?

Note that most carbonates behave this way under heat:

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