Question

What volume of 3.00 MM HClHCl in liters is needed to react completely (with nothing left over) with 0.750 LL of 0.500 MM Na2CO3Na2CO3?

Answer 1

Answer: The volume of HCl needed is 0.250 L

Explanation:

To calculate the number of moles for given molarity, we use the equation:

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

For sodium carbonate:

Molarity of sodium carbonate solution = 0.500 M

Volume of solution = 0.750 L

Putting values in above equation, we get:

$0.500M=\frac{\text{Moles of sodium carbonate}}{0.750}\\\\\text{Moles of sodium carbonate}=(0.500mol/L\times 0.750L)=0.375mol$

The chemical equation for the reaction of sodium carbonate and HCl follows:

$Na_2CO_3+2HCl\rightarrow 2NaCl+H_2CO_3$

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 = $\frac{2}{1}\times 0.375=0.750mol$ 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:

$3.00M=\frac{0.750mol}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.750mol}{3.00mol/L}=0.250L$

Hence, the volume of HCl needed is 0.250 L