Question

If 18.5 grams of CuCl2 react with 22.8 grams of NaNO3, what mass of NaCl can be formed?

Answer 1

Answer:- 15.7 grams of NaCl can be formed.

Solution:- The balanced equation is:

$CuCl_2+2NaNO_3\rightarrow Cu(NO_3)_2+2NaCl$

From this balanced equation, there is 1:2 mol ratio between $CuCl_2$ and $NaNO_3$ .

Let's calculate the grams of NaCl for the given grams of both the reactants using stoichiometry and see which one gives least amount of it. The least amount of NaCl formed would be the answer to our problem.

The calculations are as follows:

$18.5g CuCl_2(\frac{1mol CuCl_2}{134.45g CuCl_2})(\frac{2mol NaCl}{1mol CuCl_2})(\frac{58.44g NaCl}{1mol NaCl})$

= 16.1 g NaCl

$22.8g NaNO_3(\frac{1mol NaNO_3}{84.99g NaNO_3})(\frac{2mol NaCl}{2mol NaNO_3})(\frac{58.44g NaCl}{1mol NaCl})$

= 15.7 g NaCl

From above calculations, sodium nitrate is limiting as it gives least amount of NaCl.

So, 15.7 grams of NaCl can be formed from given grams of the reactants.