Question

When 7.68 g of zinc react with hydrochloric acid, what volume of hydrogen gas will be collected at 20.00C and 740 mm Hg?

Answer 1

Answer:

V ≈ 2.9 L H₂

General Formulas and Concepts:

Chemistry - Atomic Structure

• Reading a Periodic Table
• Using Dimensional Analysis

Chemistry - Reactions

• Aqueous Solutions and states of matter
• Reaction Prediction

Chemistry - Gas Laws

Combined Gas Law: PV = nRT

• P is pressure
• V is volume in liters
• n is amount of moles of substance
• R is a constant - 62.4 (L · mmHg)/(mol · K)
• T is temperature in Kelvins

Temperature Conversion: K = °C + 273.15

Explanation:

Step 1: Define

Unbalanced RxN:   Zn (s) + HCl (aq) → ZnCl₂ (aq) + H₂ (g)

Balanced RxN:   Zn (s) + 2HCl (aq) → ZnCl₂ (aq) + H₂ (g)

Given:   7.68 g Zn, 20.00 °C, 740 mmHg

Step 2: Identify Conversions

Kelvin Conversion

Molar Mass of Zn - 65.39 g/mol

Step 3: Convert

Stoichiometry:   $7.68 \ g \ Zn(\frac{1 \ mol \ Zn}{65.39 \ g \ Zn} )(\frac{1 \ mol \ H_2}{1 \ mol \ Zn} )$ = 0.117449 mol H₂

Temp Conversion:   20.00 + 273.15 = 293.15 K

Step 4: Find V

1. Substitute:                                                                                                    (740 mmHg)V = (0.117449 mol)(62.4 (L · mmHg)/(mol · K))(293.15 K)
2. Multiply:                                                                                                        (740 mmHg)V = 2148.45 L · mmHg
3. Isolate V:                                                                                                           V = 2.9033 L H₂

Step 5: Check

We are given 2 sig figs as our lowest. Follow sig fig rules and round.

2.9033 L H₂ ≈ 2.9 L H₂