Question

Consider the chemical equation. CuCl2 + 2NaNO3 mc023-1.jpg Cu(NO3)2 + 2NaCl What is the percent yield of NaCl if 31.0 g of CuCl2 reacts with excess NaNO3 to produce 21.2 g of NaCl?

Answer 1

Answer:

The answer is 78.7%

Explanation:

I just took the quiz on E2020.

Answer 2

CuCl2 + 2NaNO3 ---->  Cu(NO3)2 + 2NaCl

using molar masses:-
Theoretical yields:-
63.54 + 2(35.45) g  of CuCl2  produces  2(22.98 + 35.45) g of NaCl
    134.44  g .................................................... 116.86 g
       31.0 g ....................................................31.0 * 116.86 /134.44=26.95g  
 
 So percentage yield is 21.2* 100 / 26.95    =  78.7%  to nearest tenth