Questions: Given the unbalanced equation - Ca(OH)2 +-(NH4)2 SO4 -...+ CaSO4 +... NH3 + H2O What is the sum of all coefficients when the equation is correctly balanced using the smallest whole-number coefficients? (A) 5 (B) 7 (C) 9 (D) 11

Given the unbalanced equation - Ca(OH)2 +-(NH4)2 SO4 -...+ CaSO4 +... NH3 + H2O What is the sum of all coefficients when the equation is correctly balanced using the smallest whole-number coefficients? (A) 5 (B) 7 (C) 9 (D) 11
Transcript text: Given the unbalanced equation \[ -\mathrm{Ca}(\mathrm{OH})_{2}+-\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}-\cdots+\mathrm{CaSO}_{4}+\ldots \mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{O} \] What is the sum of all coefficients when the equation is correctly balanced using the smallest whole-number coefficients? (A) 5 (B) 7 (C) 9 (D) 11
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Solution

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Solution Steps

Step 1: Write the Unbalanced Equation

The given unbalanced equation is: \[ \mathrm{Ca(OH)}_2 + \left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4 \rightarrow \mathrm{CaSO}_4 + \mathrm{NH}_3 + \mathrm{H}_2 \mathrm{O} \]

Step 2: Identify the Atoms to Balance

We need to balance the following atoms: Ca, O, H, N, S.

Step 3: Balance Calcium (Ca)

There is 1 Ca atom on both sides of the equation: \[ \mathrm{Ca(OH)}_2 + \left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4 \rightarrow \mathrm{CaSO}_4 + \mathrm{NH}_3 + \mathrm{H}_2 \mathrm{O} \]

Step 4: Balance Sulfur (S)

There is 1 S atom on both sides of the equation: \[ \mathrm{Ca(OH)}_2 + \left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4 \rightarrow \mathrm{CaSO}_4 + \mathrm{NH}_3 + \mathrm{H}_2 \mathrm{O} \]

Step 5: Balance Nitrogen (N)

There are 2 N atoms in \(\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4\) and 1 N atom in \(\mathrm{NH}_3\). Therefore, we need 2 \(\mathrm{NH}_3\) molecules: \[ \mathrm{Ca(OH)}_2 + \left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4 \rightarrow \mathrm{CaSO}_4 + 2\mathrm{NH}_3 + \mathrm{H}_2 \mathrm{O} \]

Step 6: Balance Hydrogen (H)

There are 8 H atoms in \(\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4\) and 2 H atoms in \(\mathrm{Ca(OH)}_2\), making a total of 10 H atoms on the reactant side. On the product side, there are 6 H atoms in 2 \(\mathrm{NH}_3\) and 2 H atoms in \(\mathrm{H}_2 \mathrm{O}\), making a total of 8 H atoms. To balance, we need 2 more H atoms, so we add another \(\mathrm{H}_2 \mathrm{O}\): \[ \mathrm{Ca(OH)}_2 + \left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4 \rightarrow \mathrm{CaSO}_4 + 2\mathrm{NH}_3 + 2\mathrm{H}_2 \mathrm{O} \]

Step 7: Balance Oxygen (O)

There are 2 O atoms in \(\mathrm{Ca(OH)}_2\) and 4 O atoms in \(\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4\), making a total of 6 O atoms on the reactant side. On the product side, there are 4 O atoms in \(\mathrm{CaSO}_4\) and 2 O atoms in 2 \(\mathrm{H}_2 \mathrm{O}\), making a total of 6 O atoms. The equation is now balanced.

Final Answer

The balanced equation is: \[ \mathrm{Ca(OH)}_2 + \left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4 \rightarrow \mathrm{CaSO}_4 + 2\mathrm{NH}_3 + 2\mathrm{H}_2 \mathrm{O} \] The sum of all coefficients is: \[ 1 + 1 + 1 + 2 + 2 = 7 \] \[ \boxed{7} \] The answer is (B) 7.

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