Question

B) calculate the ratio of moles of h2o to moles of anhydrous kal(so4)2. note: report the ratio to the closest whole number. (show all work including units)

Answer 1

Answer: -

Mass of Hydrated KAl(SO₄)₂ = 2.0 g

Molar mass of anhydrous KAl(SO₄)₂ = 258.20 g/ mol

Mass of of anhydrous KAl(SO₄)₂ = mass of the 2nd heating = A

The mass of water released = mass of the Aluminum Cup + 2.0 grams of KAl(SO₄)₂ - mass of the 2nd heating

= H g

Moles of water released = $\frac{H g}{18 g/mol}$

Moles of anhydrous KAl(SO₄)₂ = $\frac{A g}{258.20 g/mol}$

Required ratio = $\frac{moles of water}{moles of anhydrous KAl(SO4)2}$

Answer 2

Moles are defined as the base unit of an element, such that 1 mole is equivalent to 6.022 $\times 10^{23}$ particles. In the given question, ratio of moles of H$_2$O to the moles of anhydrous KAl(SO$_4$)$_2$ is 12:1.

Given that,

• Chemical formula of compound Aluminum potassium sulfate = KAl(SO$_4$)$_2$
• Mass of KAl(SO$_4$)$_2$ = 258.21 g/mol
• Mass of anhydrous aluminum potassium sulfate = [mass of aluminum cup + alum after 2nd heating] –[ mass of empty cup]  
• Mass of anhydrous aluminum potassium sulfate = 3.5 g – 2.4 g  = 1.1 g

Now,

• Moles = $\dfrac {\text {Given mass}}{\text{Molar mass}}$

• $\begin{aligned}\text{Moles}&=\dfrac{1.1 \text g} {258.21 \text{g/mol}}\\\\&= 0.00426 \;\text{moles}\end{aligned}$

Now, the ratio of H$_2$O to KAl(SO$_4$)$_2$ is:

• $\begin{aligned}\text{Moles}&= \dfrac{\text{Given mass}}{\text{Molar Mass}}\\\\\text{Moles}&= \dfrac{0.05\; \text{mol H}_2 {\text O}}{0.00462\; \text{mol KAl (SO}{_4)}}_{2}} \\\\&=11.7\end$

The 11.7 to its closest whole number is 12.

Therefore, the required ratio between water and aluminum potassium sulphate is 12:1.

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