Question

A $5.00\times {10}^2$ mL sample of $2.0$ M HCl is treated with $4.47$ g of magnesium. The concentration of the acid solution after all the metal has reacted is $y\times {10}^{-3}$ M. Assume that the volume remains unchanged. Here $y$ is :

Answer 1

The balanced chemical equation is $Mg+2HCl\to MgC{l}_2+H_2$.
The number of moles of magnesium are $\frac{4.47}{24.3}=0.184$ moles.
The number of moles of $HCl$ are $2.00$ M $\times \frac{5.00\times {10}^{2}mL}{1.00\times {10}^{3}mL}=1.00$ moles.
The number of moles of $HCl$ that react with $0.184$ moles of Mg are $2\times 0.184=0.368$ moles.
The number of unreacted moles of HCl is $1.00-0.368=0.632$ moles.
The concentration of the acid solution after the metal has reacted is $\frac{0.632}{\frac{5.00\times {10}^2}{1.00\times {10}^3}}=1.264$ M $=1264\times {10}^{-3}$ M.