Question

How many grams of zinc chloride could be formed from the reaction of 3.57g of zinc with excess HCl?

Answer 1

mZn: 65 g/mol
mZnCl₂: 65g+35,5g×2 = 136 g/mol
.....................
Zn + 2HCl -----> ZnCl₂ + H₂
65g.......................136g............

65g Zn ---- 136g ZnCl₂
3,57g Zn ---- X
X = (3,57×136)/65
X = 7,47g ZnCl₂

Answer 2

Answer: 6.8 grams of zinc chloride

Explanation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

$\text{Number of moles of zinc}=\frac{3.57g}{65g/mol}=0.05moles$

$Zn(s)+2HCl(aq)\rightarrow ZnCl_2(aq)+H_2(g)$

$Zn$ is a limiting reagent as it limits the formation of products and $HCl$ is an excess reagent.

According to stoichiometry:

As 1 mole of $Zn$ gives= 1 mole of $ZnCl_2$

0.05 moles of $Zn$ will give=$\frac{1}{1}\times 0.05=0.05moles$ of $ZnCl_2$

Mass of $ZnCl_2=moles\times {\text {molar mass}}=0.05\times 136=6.8g$

Thus 6.8 grams of zinc chloride could be formed from the reaction of 3.57g of zinc with excess HCl.