Question

If 18.5 grams of CuCl2 react with 22.8 grams of NaNO3, what mass of NaCl can be formed? CuCl2 + NaNO3 → Cu(NO3)2 + NaCl

Answer 1

Answer:- 15.7 g NaCl

Solution:- It is stoichiometry problem. The balanced equation is:

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

Let's calculate the moles of NaCl from given masses of both the reactants and see which one gives least amount of it.

The calculations are as shown below:

Calculations for the grams of NaCl from $CuCl_2$ :

$18.5gCuCl_2(\frac{1moleCuCl_2}{134.45gCuCl_2})(\frac{2moleNaCl}{1moleCuCl_2})(\frac{58.44gNaCl}{1moleNaCl})$

= 16.1 g NaCl

Calculations for the grams of NaCl from $NaNO_3$ :

$22.8gNaNO_3(\frac{1moleNaNO_3}{84.99gNaNO_3})(\frac{2moleNaCl}{2moleNaNO_3})(\frac{58.44gNaCl}{1moleNaCl})$

= 15.7 g NaCl

From above calculations, $NaNO_3$ gives the limited amount of NaCl.

So, 15.7 g of NaCl would form from given amounts of reactants.

Answer 2

Answer:

B. 15.7g NaCl

Explanation:

I took the exam and got it correct