Answer :
To determine the maximum volume of ammonia gas ([tex]\( NH_3 \)[/tex]) that can be formed from heating ammonium chloride ([tex]\( NH_4Cl \)[/tex]) with excess calcium hydroxide, we need to follow these steps:
1. Calculate the molar mass of ammonium chloride ([tex]\( NH_4Cl \)[/tex]):
Ammonium chloride is composed of:
- Nitrogen (N): 1 atom with atomic mass of 14.0
- Hydrogen (H): 4 atoms with atomic mass of 1.00 each
- Chlorine (Cl): 1 atom with atomic mass of 35.45
The molar mass of [tex]\( NH_4Cl \)[/tex] can be calculated as:
[tex]\[ \text{Molar mass of } NH_4Cl = (14.0) + (4 \times 1.00) + (35.45) = 53.45 \, \text{g/mol} \][/tex]
2. Calculate the moles of [tex]\( NH_4Cl \)[/tex] used:
Given that 10.0 grams of [tex]\( NH_4Cl \)[/tex] is used, we can find the number of moles by:
[tex]\[ \text{Moles of } NH_4Cl = \frac{10.0 \, \text{g}}{53.45 \, \text{g/mol}} \approx 0.187 \, \text{moles} \][/tex]
3. Determine the moles of [tex]\( NH_3 \)[/tex] produced:
According to the balanced chemical equation:
[tex]\[ 2 \, NH_4Cl (s) + Ca(OH)_2 (s) \rightarrow CaCl_2 (s) + 2 \, NH_3 (g) + 2 \, H_2O (l) \][/tex]
The stoichiometry shows that 2 moles of [tex]\( NH_4Cl \)[/tex] produce 2 moles of [tex]\( NH_3 \)[/tex]. This is a 1:1 molar ratio. Therefore, the moles of [tex]\( NH_3 \)[/tex] produced will be equal to the moles of [tex]\( NH_4Cl \)[/tex] used:
[tex]\[ \text{Moles of } NH_3 = 0.187 \, \text{moles} \][/tex]
4. Calculate the volume of [tex]\( NH_3 \)[/tex] gas produced at room temperature and pressure:
At room temperature and pressure (RTP), 1 mole of any gas occupies 24 dm³. Therefore:
[tex]\[ \text{Volume of } NH_3 \text{ gas} = \text{Moles of } NH_3 \times 24 \, \text{dm}^3/\text{mol} \][/tex]
Substituting the value we have:
[tex]\[ \text{Volume of } NH_3 \text{ gas} = 0.187 \, \text{moles} \times 24 \, \text{dm}^3/\text{mol} \approx 4.49 \, \text{dm}^3 \][/tex]
Therefore, the maximum volume of ammonia gas ([tex]\( NH_3 \)[/tex]) that could be formed is [tex]\( 4.49 \, \text{dm}^3 \)[/tex].
1. Calculate the molar mass of ammonium chloride ([tex]\( NH_4Cl \)[/tex]):
Ammonium chloride is composed of:
- Nitrogen (N): 1 atom with atomic mass of 14.0
- Hydrogen (H): 4 atoms with atomic mass of 1.00 each
- Chlorine (Cl): 1 atom with atomic mass of 35.45
The molar mass of [tex]\( NH_4Cl \)[/tex] can be calculated as:
[tex]\[ \text{Molar mass of } NH_4Cl = (14.0) + (4 \times 1.00) + (35.45) = 53.45 \, \text{g/mol} \][/tex]
2. Calculate the moles of [tex]\( NH_4Cl \)[/tex] used:
Given that 10.0 grams of [tex]\( NH_4Cl \)[/tex] is used, we can find the number of moles by:
[tex]\[ \text{Moles of } NH_4Cl = \frac{10.0 \, \text{g}}{53.45 \, \text{g/mol}} \approx 0.187 \, \text{moles} \][/tex]
3. Determine the moles of [tex]\( NH_3 \)[/tex] produced:
According to the balanced chemical equation:
[tex]\[ 2 \, NH_4Cl (s) + Ca(OH)_2 (s) \rightarrow CaCl_2 (s) + 2 \, NH_3 (g) + 2 \, H_2O (l) \][/tex]
The stoichiometry shows that 2 moles of [tex]\( NH_4Cl \)[/tex] produce 2 moles of [tex]\( NH_3 \)[/tex]. This is a 1:1 molar ratio. Therefore, the moles of [tex]\( NH_3 \)[/tex] produced will be equal to the moles of [tex]\( NH_4Cl \)[/tex] used:
[tex]\[ \text{Moles of } NH_3 = 0.187 \, \text{moles} \][/tex]
4. Calculate the volume of [tex]\( NH_3 \)[/tex] gas produced at room temperature and pressure:
At room temperature and pressure (RTP), 1 mole of any gas occupies 24 dm³. Therefore:
[tex]\[ \text{Volume of } NH_3 \text{ gas} = \text{Moles of } NH_3 \times 24 \, \text{dm}^3/\text{mol} \][/tex]
Substituting the value we have:
[tex]\[ \text{Volume of } NH_3 \text{ gas} = 0.187 \, \text{moles} \times 24 \, \text{dm}^3/\text{mol} \approx 4.49 \, \text{dm}^3 \][/tex]
Therefore, the maximum volume of ammonia gas ([tex]\( NH_3 \)[/tex]) that could be formed is [tex]\( 4.49 \, \text{dm}^3 \)[/tex].
