Question

how many moles of aluminum are needed to produce 0.418 mol of Al2(SO4)3? 2 Al(s) + 3 H2SO4(aq) → Al2(SO4)3(aq) + 3 H2(g)

Answer 1

Answer:

0.836 mol Al

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right

Chemistry

Stoichiometry

• Using Dimensional Analysis
• Reactions RxN

Explanation:

Step 1: Define

[RxN - Balanced] 2Al (s) + 3H₂SO₄ (aq) → Al₂(SO₄)₃ (aq) + 3H₂ (g)

[Given] 0.418 mol Al₂(SO₄)₃

[Solve] x mol Al

Step 2: Identify Conversions

[RxN] 2 mol Al (s) → 1 mol Al₂(SO₄)₃ (aq)

Step 3: Stoich

1. [DA] Set up:                                                                                                      $\displaystyle 0.418 \ mol \ Al_2(SO_4)_3(\frac{2 \ mol \ Al}{1 \ mol \ Al_2(SO_4)_3})$
2. [DA] Multiply/Divide [Cancel out units]:                                                            $\displaystyle 0.836 \ mol \ Al$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

Since our final answer already has 3 sig figs, there is no need to round.