Q148CP - Chemistry: Molecular Nature Of Matter And Change

Question

Alum [KAI(SO₄)₂ .xH₂O] is used in food preparation, dye fixation, and water purification. To prepare alum, aluminum is reacted with potassium hydroxide and the product with sulfuric acid. Upon cooling, alum crystallizes from the solution. (a) A 0.5404-g sample of alum is heated to drive off the waters of hydration, and the resulting KAI(SO₄)₂ weighs 0.2941 g. Determine the value of and the complete formula of alum. (b) When 0.7500 g of aluminum is used, 8.500 g of alum forms. What is the percent yield?

Step-by-Step Solution

Answer

1. The value of x is 12 and the complete formula of alum is [KAI(SO₄)₂ .12H₂O]
2. The % yield of AI is 64.39 %.

Step 1: Determine the mass of water

The mass of alum with water [KAI(SO₄)₂ .xH₂O] = 0.5404 g

The mass of alum without water [KAI(SO₄)₂] = 0.2941 g

Thus, the mass of water is calculated as:

$\begin{array}{rcl}mass of H_{2}O& =& mass of alum with water-mass of alum without water\\ & =& 0.5404g-0.2941g\\ & =& 0.2463g\end{array}$

Step 2: Calculate the number of moles of H 2 O

The mass of H₂O = 0.2463 g

The molar mass of H₂O = 18.02 g/mol

The number of moles of H₂O is calculated as:

$\begin{array}{rcl}moles of H_{2}O& =& \frac{mass of H_{2}O}{Molar mass of H_{2}O}\\ & =& \frac{0.2463g}{18.02g/mol}\\ & =& 0.0137mol\end{array}$

Step 3: Calculate the number of moles of alum without water

The molar mass alum without water [kai(SO₄)₂] = 258.21 g/mol

The number of moles of alum without water [kai(SO₄)₂] is calculated as:

$\begin{array}{rcl}Number of moles of\left[KAI{\left(S{O}_4\right)}_2\right]& =& \frac{MASS OF [KAI(SO}{}\end{array}$₄)₂]Molar mass of [KAI(SO₄)₂] =0.2941g258.21g/mol=0.0011mol

Step 4: Calculate the number of H 2 O molecules

The number of H₂O molecules present in alum is determined by the ratio of the number of moles of alum without water and the number of moles of H₂O.

Thus, the number of H₂O molecules is calculated as:

$\begin{array}{rcl}molecules of H_{2}O& =& \frac{Moles of \left[KAI{\left(S{O}_4\right)}_2\right]}{moles of H_{2}O}\\ & =& \frac{0.0137mol}{0.0011mol}\\ & =& 12.45\end{array}$

Therefore the number of molecules of H₂O is 12.

Hence, the value of X is 12, and the complete formula of alum is [KAI(SO₄)₂ .12H₂O] .

Step 5: Calculate the number of moles of alum formed

The mass of alum formed = 8.500 g

The molar mass of alum formed = 474.35 g/mol

The number of moles of alum formed is calculated as:

$\begin{array}{rcl}Number of moles of alumformed& =& \frac{mass of alumformed}{Molar mass of alumformed}\\ & =& \frac{8.500g}{474.35g/mol}\\ & =& 0.0179mol\end{array}$

Step 6: Calculate the number of moles of aluminum used in the formation of alum

The formation reaction of alum is:

$2AI\left(s\right)+2KOH\left(aq\right)+22H_{2}O\left(l\right)+4H_{2}S{O}_4\left(aq\right)\to 2\left[KAI{\left(S{O}_4\right)}_2. 12H_{2}O\right]\left(s\right)+3H_2\left(g\right)$

Since 2 mol of  forms 2 mol of [KAI(SO₄)₂. 12H₂O] So, the number of moles of AI is:

$\begin{array}{rcl}2mol of\left[KAI{\left(S{O}_4\right)}_2 .12H_{2}O\right]formed& =& 2mol of AI\\ 0.0179molof\left[KAI{\left(S{O}_4\right)}_2 .12H_{2}O\right]formed& =& 0.0179\times \frac{2}{2}mol of AI\\ & =& 0.0179mol of AI\end{array}$

Step 7: Calculate the mass of aluminum used in the formation of alum

The molar mass of AI = 26.98 g/mol

The mass of AI is calculated as:

$\begin{array}{rcl}mass of AI& =& moles of AI\times Molar mass of AI\\ & =& 0.0179mol\times 26.98g/mol\\ & =& 0.4829g\end{array}$

Step 8: Calculate the % yield of aluminum

The % yield of AI is calculated as:

$\begin{array}{rcl}\%yield of AI& =& \frac{actualyield}{Theoreticalyield}\times 100\%\\ & =& \frac{0.4829g}{0.7500g}\times 100\%\\ & =& 64.39\%\end{array}$