# How many grams of CaO can be formed from the decomposition of 275 g of CaCO_3?

Jan 7, 2017

Approx. $150 \cdot g$ $\text{calcium oxide}$ will be isolated given stoichiometric reaction.

#### Explanation:

We need (i) a stoichiometric equation:

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

And (ii) equivalent quantities of $\text{calcium carbonate}$,

$\text{Moles of calcium carbonate} = \frac{275 \cdot g}{100.09 \cdot g \cdot m o {l}^{-} 1} = 2.75 \cdot m o l$

And thus 2.75*molxx56.08*g*mol^-1=??*g $\text{calcium oxide}$.

Why is there $\text{1:1}$ equivalence between $\text{calcium oxide}$ and $\text{calcium carbonate}$.