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How much ZnCl₂ can be produced from the reaction of 2.5 g of Zn and 5.0 g of HCl?

A) 1.75 g
B) 3.25 g
C) 4.50 g
D) 6.75 g

User Dykam
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1 Answer

1 vote

Final answer:

The amount of ZnCl₂ produced from the reaction can be determined using stoichiometry. The moles of Zn and HCl are calculated, and since 1 mole of Zn produces 1 mole of ZnCl₂, the moles of ZnCl₂ produced will be equal to the moles of Zn. Therefore, the amount of ZnCl₂ produced is 0.038 mol.

Step-by-step explanation:

To determine the amount of ZnCl₂ produced from the reaction, we need to use stoichiometry. First, calculate the moles of Zn and HCl:

Moles of Zn = mass of Zn / molar mass of Zn = 2.5 g / 65.38 g/mol = 0.038 mol

Moles of HCl = mass of HCl / molar mass of HCl = 5.0 g / 36.46 g/mol = 0.137 mol

According to the balanced equation Zn + 2HCl → ZnCl₂ + H₂, the ratio between Zn and ZnCl₂ is 1:1. This means that for every 1 mole of Zn, 1 mole of ZnCl₂ is produced.

Therefore, the moles of ZnCl₂ produced will be equal to the moles of Zn.

So, the amount of ZnCl₂ produced is 0.038 mol.

User SHASHI BHUSAN
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