Your idea matters-17

The reaction looks like this:

$\ce{CaCO3 + \text{heat} -> CaO + CO2}$

First, we need to calculate the # of moles of $\ce{CaCO3}$ in $15 g \text{ of } \ce{CaCO3}$

Molar Mass of $\ce{CaCO3}$ $= 100.0869 \text{ g/mol}$

Using Dimensional Analysis…

$\begin{array}{c|c} \text{ 15 g } \ce{CaCO3} & \text{ 1 mol} \ce{CaCO3} \\ \hline & \text{ 100.0869 g } \ce{CaCO3} \end{array}$ $= 0.150 \text{ mol } \ce{CaCO3}$

So now that we know that $0.150 \text{ mol } \ce{CaCO3}$ goes into the reaction, we need to find how much of that amount contributes to the resulting $\ce{CaO}$ .

Molar Mass of $\ce{CaO}$ $= 56.0774 \text{ g/mol}$

So, again, using Dimensional Analysis…

$\begin{array}{c|c} \text{ 0.150 mol } \ce{CaCO3} & \text{ 56.0774 g } \ce{CaO} \\ \hline & \text{ 1 mol } \ce{CaO} \end{array}$ $= 8.40 \text{ g } \ce{CaO}$