Question

Sodium carbonate reacts with nitric acid to form sodium nitrate, carbon dioxide, and water according to the balanced reaction below. How many grams of sodium carbonate are required to completely react with 25.0 mL of 0.155 M nitric acid?

Answer 1

Answer: The mass of sodium carbonate reacted is 0.205 g.

Explanation:

To calculate the moles of a solute, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

We are given:

Volume of nitric acid = 25 mL = 0.025 L   (Conversion factor: 1 L = 1000 mL)

Molarity of the solution = 0.155 moles/ L

Putting values in above equation, we get:

$0.155mol/L=\frac{\text{Moles of nitric acid}}{0.025L}\\\\\text{Moles of nitric acid}=0.003875mol$

For the given chemical reaction:

$Na_2CO_3+2HNO_3\rightarrow H_2CO_3+2NaNO_3$

By Stoichiometry of the reaction:

If 2 moles of nitric acid reacts with 1 mole of sodium carbonate.

So, 0.003875 moles of nitric acid will react with $\frac{1}{2}\times 0.003875=0.0019375moles$ of sodium carbonate.

To calculate the mass of sodium carbonate, we use the equation:

Molar mass of sodium carbonate = 105.98 g/mol

Moles of sodium carbonate = 0.0019375 moles

Putting values in above equation, we get:

$0.0019375mol=\frac{\text{Mass of sodium carbonate}}{105.98g/mol}\\\\\text{Mass of sodium carbonate}=0.205g$

Hence, the mass of sodium carbonate reacted is 0.205 g.