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

The answer is 78.7%

Step-by-step 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