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)

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

The balanced chemical equation is

The conversions are  

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

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

Moles to Molarity of

(Answer)

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