Answer: The volume of HCl needed is 0.250 L
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
For sodium carbonate:
Molarity of sodium carbonate solution = 0.500 M
Volume of solution = 0.750 L
Putting values in above equation, we get:
[tex]0.500M=\frac{\text{Moles of sodium carbonate}}{0.750}\\\\\text{Moles of sodium carbonate}=(0.500mol/L\times 0.750L)=0.375mol[/tex]
The chemical equation for the reaction of sodium carbonate and HCl follows:
[tex]Na_2CO_3+2HCl\rightarrow 2NaCl+H_2CO_3[/tex]
By Stoichiometry of the reaction:
1 mole of sodium carbonate reacts with 2 moles of HCl
So, 0.375 moles of sodium carbonate will react with = [tex]\frac{2}{1}\times 0.375=0.750mol[/tex] of HCl
Now, calculating the volume of HCl by using equation 1:
Moles of HCl = 0.750 moles
Molarity of HCl = 3.00 M
Putting values in equation 1, we get:
[tex]3.00M=\frac{0.750mol}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.750mol}{3.00mol/L}=0.250L[/tex]
Hence, the volume of HCl needed is 0.250 L
