Zn+CuCl2→ZnCl2+Cu

How many grams of ZnCl2 will be produced from 7.00 grams of Zn and 8.00 grams of CuCl2?

The answer to your question is 8.1 g of ZnCl₂

Explanation:

Data

mass of ZnCl₂ = ?

mass of Zn = 7.0 g

mass of CuCl₂ = 8 g

Balanced chemical reaction

                Zn  +  CuCl₂   ⇒   ZnCl₂  +  Cu

Process

1.- Calculate the molar mass of the reactants

Zn = 65.4 g

CuCl₂ = 63.6 + (35.5 x 2) = 134.6 g

2.- Calculate the limiting reactant

theoretical yield  CuCl₂ / Zn = 134.6/65.4 = 2.06

experimental yield CuCl₂ / Zn = 8/7 = 1.1

The limiting reactant is CuCl₂ because the experimental yield was lower than the theoretical yield.

3.- Use proportions to calculate the mass of ZnCl₂

molar mass of ZnCl₂ = 65.4 + (35.5 x 2) = 136.4 g

                  134.6 g of CuCl₂ 136.4 g of ZnCl₂

                      8   g                  x

                      x = (8 x 136.4) / 134.6

                      x = 8.1 g of ZnCl₂

oh it is true if thats what you are asking

the dependant variable is spilled water

the independent is the heat