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galina1969 [7]
4 years ago
8

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

Chemistry
2 answers:
anyanavicka [17]4 years ago
7 0

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.

lorasvet [3.4K]4 years ago
5 0
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
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1 year ago
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Answer:

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Explanation:

Let's clarify the irrelevant terms first.

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3 years ago
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