Ammonium chloride reacts with calcium hydroxide according to the following balanced equation:

[tex]\[ 2 \text{NH}_4 \text{Cl} (s) + \text{Ca(OH)}_2 (s) \rightarrow 2 \text{NH}_3 (g) + \text{CaCl}_2 (s) + 2 \text{H}_2 \text{O} (l) \][/tex]

If [tex]\(5.60 \, \text{g}\)[/tex] of [tex]\(\text{Ca(OH)}_2\)[/tex] is used up completely in this reaction, calculate:

1. The number of moles of [tex]\(\text{NH}_4 \text{Cl}\)[/tex] required for the reaction.



Answer :

Sure, let's go through the steps to solve the problem.

1. Identify the given data:
Mass of calcium hydroxide [tex]\((Ca(OH)_2)\)[/tex]: [tex]\(5.60 \text{ grams}\)[/tex]
Molar mass of [tex]\(Ca(OH)_2\)[/tex]: [tex]\(74.09 \text{ g/mol}\)[/tex]

2. Recall the balanced chemical equation:
[tex]\[ 2 NH_4Cl(s) + Ca(OH)_2(s) \rightarrow 2 NH_3(g) + CaCl_2(s) + 2 H_2O(l) \][/tex]

3. Calculate the number of moles of [tex]\(Ca(OH)_2\)[/tex]:
The formula to calculate the number of moles [tex]\(n\)[/tex] is:
[tex]\[ n = \frac{\text{mass}}{\text{molar mass}} \][/tex]
Given the mass of [tex]\(Ca(OH)_2\)[/tex] is [tex]\(5.60 \text{ g}\)[/tex] and its molar mass is [tex]\(74.09 \text{ g/mol}\)[/tex], we calculate:
[tex]\[ n(\text{Ca(OH)}_2) = \frac{5.60 \text{ g}}{74.09 \text{ g/mol}} = 0.0755837494938588 \text{ moles} \][/tex]

4. Stoichiometric relationship:
From the balanced chemical equation, we see that 1 mole of [tex]\(Ca(OH)_2\)[/tex] reacts with 2 moles of [tex]\(NH_4Cl\)[/tex]. Therefore the stoichiometric ratio of [tex]\(Ca(OH)_2\)[/tex] to [tex]\(NH_4Cl\)[/tex] is [tex]\(1 : 2\)[/tex].

5. Calculate the number of moles of [tex]\(NH_4Cl\)[/tex] required:
Using the stoichiometric ratio, the number of moles of [tex]\(NH_4Cl\)[/tex] required is:
[tex]\[ n(\text{NH}_4\text{Cl}) = 2 \times n(\text{Ca(OH)}_2) \][/tex]
Substituting [tex]\(n(\text{Ca(OH)}_2)\)[/tex] from step 3:
[tex]\[ n(\text{NH}_4\text{Cl}) = 2 \times 0.0755837494938588 = 0.1511674989877176 \text{ moles} \][/tex]

Therefore, 0.1511674989877176 moles of [tex]\( NH_4Cl \)[/tex] are required for the reaction.