Use the following reaction: CuCl2 + 2 NaNO3 → Cu(NO3)2 + 2 NaCl

If 90.0 grams of copper (II) chloride react with 120.0 grams of sodium nitrate, how much sodium chloride can be formed (in grams)?

78.4g

Explanation:

Given parameters:

Mass of copper (II) chloride  = 90g

Mass of sodium nitrate  = 120g

Unknown:

Mass of sodium chloride that can be formed  = ?

Solution:

The balanced chemical reaction is:

            CuCl₂   +   2NaNO₃ →   Cu(NO₃)₂   +   2NaCl

Let us find the limiting reactant. This reactant will determine the extent of the reaction of the amount of product that will be formed.

First, convert the masses to number of moles;

      Number of moles  =

Molar mass of   CuCl₂  = 63.5 + 2(35.5) = 134.5g/mol

Molar mass of NaNO₃   = 23 + 14 + 3(16) = 85g/mol

  Number of moles of CuCl₂ =  = 0.67mole

  Number of moles of NaNO₃  =  = 1.41mole

From the balanced reaction equation:

      1 mole of CuCl₂   would react with 2 moles of NaNO₃

     0.67mole of CuCl₂   would require   0.67 x 2  = 1.34mole of NaNO₃

So, CuCl₂ is the limiting reactant

     1 mole of CuCl₂ will produce 2 mole of NaCl

    0.67mole of CuCl₂ will therefore yield 0.67 x 2 = 1.34mole of NaCl

Mass of NaCl = number of moles x molar mass

  Molar mass of NaCl = 23 + 35.5 = 58.5g/mol

Mass of NaCl  = 58.5 x 1.34  = 78.4g