Question

What mass of zinc chloride can be produced by reacting 50.0 mL of 1.00 mol/L HCl solution with sufficient zinc?

a) 136 g
b) 64 g
c) 54 g
d) 27 g

Answer 1

To find the mass of zinc chloride produced from 50.0 mL of 1.00 M HCl, one must calculate the moles of HCl, use the stoichiometry of the reaction with zinc, and then find the mass of zinc chloride. The correct answer is not listed in the options; the calculated mass is approximately 3.41 g.

Step-by-step explanation:

To calculate the mass of zinc chloride produced from 50.0 mL of 1.00 mol/L HCl solution reacting with zinc, we must first understand the stoichiometry of the reaction between zinc and hydrochloric acid. The balanced chemical equation is:

Zn(s) + 2 HCl(aq) → ZnCl₂(aq) + H₂(g)

From the equation, we see that one mole of zinc reacts with two moles of HCl to produce one mole of zinc chloride. Since we have a 1.00 M HCl solution, and the volume of the solution is 50.0 mL, we can convert the volume to liters (0.050 L) and calculate the moles of HCl:

Moles of HCl = 1.00 mol/L × 0.050 L = 0.050 mol

Now, this amount of HCl will react with half the amount of zinc (0.025 mol), because it takes two moles of HCl to react with one mole of zinc. Using the molar mass of zinc chloride (136.30 g/mol), we can find the mass:

Mass of ZnCl₂ = 0.025 mol × 136.30 g/mol = 3.4075 g

The closest answer from the options provided is option d) 27 g, which is clearly not correct, as the calculated mass is much lower, approximately 3.41 g. Therefore, it appears there may be a mistake in the question or the options provided.

Answer 2

The mass of zinc chloride produced from 50.0 mL of 1.00 mol/L HCl is 3.4 g, a result that doesn't match any of the given options, suggesting an error in the question or options.

Step-by-step explanation:

The mass of zinc chloride that can be produced by reacting 50.0 mL of 1.00 mol/L HCl solution with sufficient zinc can be calculated using stoichiometry. First, we convert the volume of HCl to moles (volume × molarity), which gives us 0.050 L × 1 mol/L = 0.05 moles of HCl. The reaction follows a 2:1 stoichiometry between HCl and Zn (2 moles of HCl react with 1 mole of Zn, producing 1 mole of ZnCl₂), so we have enough HCl to fully react with 0.025 moles of Zn. Using the molar mass of ZnCl₂ (136 g/mol), the mass of ZnCl₂ produced is 0.025 moles × 136 g/mol = 3.4 g, which is not among the provided options.

In contrast, the question asked implies options for answer, suggesting that perhaps an error was made in either the question or options provided, as options (a) 136 g, (b) 64 g, (c) 54 g, and (d) 27 g do not match the calculated value.