Magnesium metal reacts with hydrochloric acid to form magnesium chloride and hydrogen gas. Suppose we react an excess of magnesium metal with 20.0 mL of a 3.00 M solution of hydrochloric acid and collect all of the hydrogen in a balloon at 25°C and 1.00 atm. What is the expected volume of the balloon?

0.734 L or 734 mL

Explanation:

We'll begin by calculating the number of mole of HCl in 20.0 mL of a 3.00 M solution of hydrochloric acid, HCl.

This is illustrated below:

Molarity of hcl = 3 M

Volume = 20 mL = 20/1000 = 0.02 L

Mole of HCl = ... ?

Molarity = mole / Volume

3 = mole / 0.02

Cross multiply

Mole of HCl = 3 x 0.02

Mole of HCl = 0.06 mole.

Next, we shall determine the number of mole of Hydrogen gas, H2 produced from the reaction of 0.06 mole of HCl. This can obtain as follow:

Mg + 2HCl - > MgCl2 + H2

From the balanced equation above,

2 moles of HCl reacted to produce 1 mole H2.

Therefore, 0.06 mole of HCl will react to produce = (0.06 x 1) / 2 = 0.03 mole

Therefore, 0.03 mole of H2 was produced.

Finally, we shall determine the volume of the balloon by calculating the volume of Hydrogen gas, H2 in the balloon.

The volume of H2 can be obtained by using the ideal gas equation as follow:

Pressure (P) = 1 atm

Temperature (T) = 25°C = 25°C + 273 = 298K

Number of mole (n) of H2 = 0.03 mole

Gas constant (R) = 0.0821 atm. L/Kmol

Volume (V) of H2 = ?

PV = nRT

1 x V = 0.03 x 0.0821 x 298

V = 0.734 L = 734 mL.

Therefore, the volume of the balloon is 0.734 L or 734 mL