Magnesium metal reacts with hydrochloric acid to form magnesium chloride and hydrogen gas. Suppose we react an excess of magnesium metal with 19.8 mL of a 3.00 M solution of hydrochloric acid and collect all of the hydrogen in a balloon at 25^oC and 1.00 atm. What is the expected volume of the balloon? A) 0.0609 L B) 0.726 L C) 1.45 L D) 0.323 L E) 60.9 L

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of hydrogen gas produced in the reaction:

Mg + 2HCl → MgCl2 + H2

1 mole of Mg reacts with 2 moles of HCl to produce 1 mole of H2. We are given the volume and concentration of the HCl solution, so we can calculate the number of moles of HCl used:

19.8 mL x (1 L / 1000 mL) x (3.00 mol / 1 L) = 0.0594 mol HCl

Because we have an excess of Mg, all of the HCl will be used up, and we can calculate the number of moles of H2 produced:

0.0594 mol HCl x (1 mol H2 / 2 mol HCl) = 0.0297 mol H2

Next, we can use the ideal gas law to calculate the volume of the balloon:

PV = nRT

P = 1.00 atm
V = ?
n = 0.0297 mol
R = 0.0821 L-atm/mol-K
T = (25 + 273) K = 298 K

V = (nRT) / P = (0.0297 mol x 0.0821 L-atm/mol-K x 298 K) / 1.00 atm = 0.726 L

Therefore, the expected volume of the balloon is 0.726 L, or choice B.

Magnesium Metal Reacts with Hydrochloric Acid to Form Magnesium Chloride