Answer:
The molarity of ZnCl2 is 1.699 mol/ L.
Explanation:
Balance chemical equation:
Zn(s) + CuCl2(aq) → ZnCl2(aq) + Cu(s)
Given data:
mass of zinc = 30 g
total volume of solution = 270 mL
molarity of ZnCl2 = ?
Solution:
First of all we will calculate the moles of Zn.
molar mass of zinc = 65.38 g/mol
number of moles = mass / molar mass
number of moles = 30 g/ 65.38 g/mol
number of moles = 0.4588 mole
From the balance chemical equation we will compare the moles of Zn and ZnCl2
Zn : ZnCl2
1 : 1
0.4588 : 0.4588
The number of moles of ZnCl2 are 0.4588 mol.
Molarity of ZnCl2:
Total volume of solution is 270 mL. We will convert it into liter.
1 L = 1000 mL
270/1000 = 0.27 L
Molarity = moles of solute / volume of solution in liter
Molarity = 0.4588 mol / 0.27 L
Molarity = 1.699 mol/ L
Answer:
The balanced chemical equation is
[tex]Zn(s)+CuCl_2 (aq)>ZnCl_2 (aq)+Cu(s)[/tex]
The conversions are
Mass Zinc to moles Zinc (dividing by molar mass of Zinc)
Moles Zinc to moles [tex]ZnCl_2[/tex] ( using mole ratio 1 : 1 )
Moles [tex]ZnCl_2[/tex] to Molarity of [tex]ZnCl_2[/tex]
[tex]moles Zn= \frac {mass}{(molar mass)} \\\\=\frac {30.0g}{(65g per mol )} = 0.462 mol[/tex]
[tex]0.462mol Zn \times \frac{(1mol ZnCl_2)}{(1mol Zn)}\\\\=0.462 mol ZnCl_2[/tex]
[tex]Molarity ZnCl_2 = \frac {(moles ZnCl_2)}{(Volume of solution in L)}[/tex]
[tex]= \frac {(0.462mol ZnCl_2)}{0.270L}=1.71 \frac {mol}{L} or M[/tex]
(Answer)
1.71 M is the molarity of [tex]ZnCl_2[/tex] that forms when 30.0 g of zinc completely reacts with [tex]CuCl_2[/tex]
