Question

Consider the following reaction: Zn(s) + 2 HCl (aq) → ZnCl₂(aq) + H₂(g) if 14.2g of Zn(s) reacts with 14.2 g of HCl, how many grams of ZnCl₂ can be produced?

a. 26.6
b. 14.4
c. 0.390
d. 0.217

Answer 1

When 14.2g of Zn reacts with 14.2g of HCl, 26.6 grams of ZnCl₂ can be produced.

Step-by-step explanation:

In the given reaction, the stoichiometry is 1 mole of Zn reacts with 2 moles of HCl to produce 1 mole of ZnCl₂ and 1 mole of H₂ gas. We need to find how many grams of ZnCl₂ can be produced when 14.2g of Zn reacts with 14.2g of HCl.

To solve this problem, we need to convert the masses of Zn and HCl to moles using their molar masses. Then, we compare the mole ratios from the balanced equation to determine the limiting reactant.

Once we identify the limiting reactant, we can use the mole ratios to calculate the mass of ZnCl₂ produced.

Using the molar masses:

• Molar mass of Zn = 65.38 g/mol
• Molar mass of HCl = 36.46 g/mol
• Molar mass of ZnCl₂ = 136.30 g/mol

We can then calculate the moles of Zn and HCl:

• Moles of Zn = 14.2 g / 65.38 g/mol = 0.217 mol
• Moles of HCl = 14.2 g / 36.46 g/mol = 0.390 mol

The moles ratio from the balanced equation is 1:2, meaning that 1 mole of Zn reacts with 2 moles of HCl. Since we have excess HCl (0.390 mol), Zn is the limiting reactant.

Therefore, the mass of ZnCl₂ produced is:

• Mass of ZnCl₂ = moles of ZnCl₂ × molar mass of ZnCl₂
• Mass of ZnCl₂ = 0.217 mol × 136.30 g/mol = 29.60 g

So, the correct answer is option b. 26.6 grams of ZnCl₂ can be produced.