Answer: The mass of sodium carbonate reacted is 0.205 g.
Explanation:
To calculate the moles of a solute, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
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:
[tex]0.155mol/L=\frac{\text{Moles of nitric acid}}{0.025L}\\\\\text{Moles of nitric acid}=0.003875mol[/tex]
For the given chemical reaction:
[tex]Na_2CO_3+2HNO_3\rightarrow H_2CO_3+2NaNO_3[/tex]
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 [tex]\frac{1}{2}\times 0.003875=0.0019375moles[/tex] 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:
[tex]0.0019375mol=\frac{\text{Mass of sodium carbonate}}{105.98g/mol}\\\\\text{Mass of sodium carbonate}=0.205g[/tex]
Hence, the mass of sodium carbonate reacted is 0.205 g.
