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.
