Question

What is the molarity of ZnCl2 that forms when 30.0 g of zinc completely reacts with CuCl2 according to the following reaction? Assume a final volume of 270 mL . Zn(s)+CuCl2(aq)→ZnCl2(aq)+Cu(s)

Answer 1

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 2

Answer:

The balanced chemical equation is

$Zn(s)+CuCl_2 (aq)>ZnCl_2 (aq)+Cu(s)$

The conversions are  

Mass Zinc to moles Zinc (dividing by molar mass of Zinc)

Moles Zinc to moles $ZnCl_2$  ( using mole ratio 1 : 1 )

Moles $ZnCl_2$ to Molarity of $ZnCl_2$

$moles Zn= \frac {mass}{(molar mass)} \\\\=\frac {30.0g}{(65g per mol )} = 0.462 mol$

$0.462mol Zn \times \frac{(1mol ZnCl_2)}{(1mol Zn)}\\\\=0.462 mol ZnCl_2$

$Molarity ZnCl_2 = \frac {(moles ZnCl_2)}{(Volume of solution in L)}$

$= \frac {(0.462mol ZnCl_2)}{0.270L}=1.71 \frac {mol}{L} or M$

(Answer)

1.71 M is the molarity of $ZnCl_2$ that forms when 30.0 g of zinc completely reacts with $CuCl_2$