Question

The answer to question 2

Answer 1

Answer:

12

Explanation:

You will need a chemical equation with masses and molar masses, so let’s gather all the information in one place.

$M_{r}$:                           258.21       18.02

                 KAl(SO₄)₂·xH₂O ⟶ KAl(SO₄)₂ + xH₂O

Mass/g:             4.74                                       2.16

Step 1. Calculate the mass of the KAl(SO₄)₂.

Mass = 4.74 g – 2.16 g = 2.58 g.

Step 2. Calculate the moles of each product.

$\text{Moles of KAl(SO}_{4})_{2} = \text{2.58 g} \times \frac{\text{1 mol} }{\text{258.21 g}} = 9.992 \times 10^{-3} \text{ mol}$

$\text{Moles of H}_{2}\text{O} = \text{2.16 g} \times \frac{\text{1 mol} }{\text{18.02 g}} = \text{ 0.1200 mol}$

Step 3. Calculate the molar ratio of the two products.

$\frac{\text{Moles of KAl(SO}_{4})_{2}}{\text{Moles of H}_{2}\text{O}} = \frac{ 9.992 \times 10^{-3} \text{ mol}}{\text{ 0.1200 mol} } = \frac{1 }{12.01} \approx \frac{ 1}{ 12}$

1 mol of KAl(SO₄)₂ combines with 12 mol H₂O, so x = 12.