Question

Reviewing for a test - need an answer and explanation.

Base your answer to the following question on the information below.
The reaction between aluminum and an aqueous solution of gold (I) sulfate is represented by the unbalanced equation below.

Al (s) + Au2SO4 (aq) -> Al2(SO4)3 (aq) + Au (s)

Determine the total mass of Au sproduced when 6.52 grams of Al reacts completely with 3.58 grams of Au2SO4 to produce 5.95 grams of Al2(SO4)3.

Answer 1

2.88 grams.

Step-by-step explanation

Al is in excess. 6.52 grams of Al converted to only 5.95 grams of Al₂(SO₄)₃. At least some Al atoms is in excess. In other words, Au₂SO₄ is the limiting reactant. All 3.58 grams of Au₂SO₄ is converted.

Refer to the conservation of atoms to avoid balancing the equation. Au atoms conserve. There are two Au atoms in one formula unit of Au₂SO₄. As a result, two moles of Au atoms will be produced when one mole formula units of Au₂SO₄ reacts completely with Al.

How many moles of formula units in 3.58 grams of Au₂SO₄?

Refer to a modern periodic table for relative atomic mass data:

• Au: 196.967;
• S: 32.06;
• O: 15.999.

`M(\text{Au}_2\text{SO}_4) = 2 * 196.967 + 32.06 + 4 * 15.999 = 489.99 \; \text{g}\cdot \text{mol}^(-1)`.

`n(\text{Au}_2\text{SO}_4) = \frac{m(\text{Au}_2\text{SO}_4)}{M(\text{Au}_2\text{SO}_4)} = \frac{3.58 \; \text{g}}{489.99\; \text{g}\cdot \text{mol}^(-1)} = 0.00731 \; \text{mol}`.

How many moles of Au atoms will be produced from 0.00731 mol of Au₂SO₄?

There are two Au atoms in each formula unit of Au₂SO₄. Both Au atoms will end up as Au (s) as Au₂SO₄ reacts with excess Al.

`n(\text{Au}) = 2 \; n(\text{Au}_2\text{SO}_4) = 2 * 0.00731 \; \text{mol} = 0.0146 \; \text{mol}`.

What's the mass of 0.0146 mol of Au?

The relative atomic mass of Au is 196.967.

`m(\text{Au}) = n(\text{Au}) \cdot M(\text{Au}) = 0.0146 * 196.967 = 2.88 \; \text{g}`.