Respuesta :
Answer: 12.0 milliliters of 6.50 M HCl ( aq ) are required to react with 2.55 g Zn.
Explanation:
moles =[tex]\frac{\text {given mass}}{\text {Molar mass}}[/tex]
moles of zinc =[tex]\frac{2.55g}{65.38g/mol}=0.0390moles[/tex]
The balanced chemical equation is :
[tex]Zn(s)+2HCl(aq)\rightarrow ZnCl_2(aq)+H_2(g)[/tex]
According to stoichiometry:
1 mole of zinc reacts with = 2 moles of HCl
Thus 0.0390 moles of zinc reacts with = [tex]\frac{2}{1}\times 0.0390=0.0780[/tex] moles of HCl
To calculate the volume for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution in ml}}[/tex] .....(1)
Molarity of [tex]HCl[/tex] solution = 6.50 M
Volume of solution = ?
Putting values in equation 1, we get:
[tex]6.50M=\frac{0.0780\times 1000}{\text{Volume of solution in ml}}[/tex]
[tex]{\text{Volume of solution in ml}}=12.0ml[/tex]
Thus 12.0 ml of 6.50 M HCl ( aq ) are required to react with 2.55 g Zn
