Question

What mass of dilute nitric acid will be required to dissolved $5\mathrm{g}$ of calcium carbonate?

Answer 1

Step 1: Given information

• Mass of calcium carbonate $\left({\mathrm{CaCO}}_3\right)=5\mathrm{g}$

Step 2: Reaction involved

• Calcium carbonate $\left({\mathrm{CaCO}}_3\right)$ reacts with nitric acid $\left({\mathrm{HNO}}_3\right)$ to form Water $\left({\mathrm{H}}_2\mathrm{O}\right)$, Carbon dioxide $\left({\mathrm{CO}}_2\right)$ and Calcium nitrate $\left(\mathrm{Ca}{\left({\mathrm{NO}}_3\right)}_2\right)$.
• The balanced reaction is as follows:

${\mathrm{CaCO}}_3\left(\mathrm{s}\right)+2{\mathrm{HNO}}_3\left(\mathrm{aq}\right)\to {\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)+\mathrm{Ca}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{aq}\right)$

Step 3: Calculate the mass of calcium carbonate and nitric acid

The atomic mass of Ca = $40\text{g}$

The atomic mass of C = $12\text{g}$

The atomic mass of O = $16\text{g}$

The atomic mass of H = $1\text{g}$

The atomic mass of N = $14\text{g}$

$\begin{array}{rcl}\mathrm{Mass}\mathrm{of}{\mathrm{CaCO}}_3& =& 40\mathrm{g}+12\mathrm{g}+16\mathrm{g}\times 3\\ & =& 100\mathrm{g}\end{array}$

$\begin{array}{rcl}\mathrm{Mass}\mathrm{of}{\mathrm{HNO}}_3& =& \left(1\mathrm{g}+14\mathrm{g}+16\mathrm{g}\times 3\right)\\ & =& 63\mathrm{g}\end{array}$

$\begin{array}{rcl}\mathrm{Mass}\mathrm{of}2\mathrm{moles}\mathrm{of}{\mathrm{HNO}}_3& =& 2\times 63\mathrm{g}\\ & =& 126\mathrm{g}\end{array}$

Step 4: Mass of dilute nitric acid

• $100\mathrm{g}$ calcium carbonate dissolved in $126\mathrm{g}$ nitric acid.
• $5\mathrm{g}$ of calcium carbonate will dissolve $\frac{126\mathrm{g}}{100\mathrm{g}}\times 5\mathrm{g}=6.3\mathrm{g}$ nitric acid.

Therefore, $6.3\mathrm{g}$ of dilute nitric acid will be required to dissolved $5\mathrm{g}$ of calcium carbonate