Question
What type of reaction is 2NH3+H2SO4 right arrow (NH4)SO4
Answers
Answer
2.7g of a gaseous hydrcarbon, which contains 11.1% of hydrogen occupies a volume of 1.12dm3 at s.t.p. what is the percentage composition of carbon in the compound
Answered by
GPT-4o mini
The reaction you provided:
\[ 2 \text{NH}_3 + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4 \]
is a type of acid-base neutralization reaction. In this reaction, ammonia (NH₃), which acts as a base, reacts with sulfuric acid (H₂SO₄), serving as an acid, to form ammonium sulfate ((NH₄)₂SO₄).
In essence, NH₃ accepts protons (H⁺ ions) from H₂SO₄, leading to the formation of the ammonium ion (NH₄⁺), resulting in the salt ammonium sulfate.
\[ 2 \text{NH}_3 + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4 \]
is a type of acid-base neutralization reaction. In this reaction, ammonia (NH₃), which acts as a base, reacts with sulfuric acid (H₂SO₄), serving as an acid, to form ammonium sulfate ((NH₄)₂SO₄).
In essence, NH₃ accepts protons (H⁺ ions) from H₂SO₄, leading to the formation of the ammonium ion (NH₄⁺), resulting in the salt ammonium sulfate.
Answered by
GPT-4o mini
To find the percentage composition of carbon in the gaseous hydrocarbon, we can follow these steps:
1. **Determine the moles of the gas using the ideal gas law**: At standard temperature and pressure (s.t.p), 1 mole of gas occupies 22.4 dm³.
2. **Find the number of moles of the gaseous hydrocarbon**:
\[
\text{Moles of gas} = \frac{\text{Volume}}{\text{Molar volume at s.t.p}} = \frac{1.12 \, \text{dm}^3}{22.4 \, \text{dm}^3/\text{mol}} \approx 0.05 \, \text{mol}
\]
3. **Determine the mass of hydrogen in the compound**:
- Given that the compound is 11.1% hydrogen by mass:
\[
\text{Mass of hydrogen} = 0.111 \times 2.7 \, \text{g} \approx 0.2997 \, \text{g}
\]
4. **Calculate the number of moles of hydrogen**:
- The molar mass of hydrogen (H) is approximately 1 g/mol:
\[
\text{Moles of H} = \frac{0.2997 \, \text{g}}{1 \, \text{g/mol}} \approx 0.2997 \, \text{mol}
\]
5. **Calculate the number of moles of carbon in the compound**:
- The total mass of the hydrocarbon is 2.7 g, so the mass of carbon (C) can be found by subtracting the mass of hydrogen:
\[
\text{Mass of carbon} = \text{Total mass} - \text{Mass of hydrogen} = 2.7 \, \text{g} - 0.2997 \, \text{g} \approx 2.4003 \, \text{g}
\]
6. **Calculate the number of moles of carbon**:
- The molar mass of carbon is approximately 12 g/mol:
\[
\text{Moles of C} = \frac{2.4003 \, \text{g}}{12 \, \text{g/mol}} \approx 0.2000 \, \text{mol}
\]
7. **Calculate the percentage composition of carbon**:
\[
\text{Percentage of C} = \left( \frac{\text{Mass of carbon}}{\text{Total mass}} \right) \times 100 = \left( \frac{2.4003 \, \text{g}}{2.7 \, \text{g}} \right) \times 100 \approx 88.89\%
\]
Thus, the percentage composition of carbon in the compound is approximately **88.89%**.
1. **Determine the moles of the gas using the ideal gas law**: At standard temperature and pressure (s.t.p), 1 mole of gas occupies 22.4 dm³.
2. **Find the number of moles of the gaseous hydrocarbon**:
\[
\text{Moles of gas} = \frac{\text{Volume}}{\text{Molar volume at s.t.p}} = \frac{1.12 \, \text{dm}^3}{22.4 \, \text{dm}^3/\text{mol}} \approx 0.05 \, \text{mol}
\]
3. **Determine the mass of hydrogen in the compound**:
- Given that the compound is 11.1% hydrogen by mass:
\[
\text{Mass of hydrogen} = 0.111 \times 2.7 \, \text{g} \approx 0.2997 \, \text{g}
\]
4. **Calculate the number of moles of hydrogen**:
- The molar mass of hydrogen (H) is approximately 1 g/mol:
\[
\text{Moles of H} = \frac{0.2997 \, \text{g}}{1 \, \text{g/mol}} \approx 0.2997 \, \text{mol}
\]
5. **Calculate the number of moles of carbon in the compound**:
- The total mass of the hydrocarbon is 2.7 g, so the mass of carbon (C) can be found by subtracting the mass of hydrogen:
\[
\text{Mass of carbon} = \text{Total mass} - \text{Mass of hydrogen} = 2.7 \, \text{g} - 0.2997 \, \text{g} \approx 2.4003 \, \text{g}
\]
6. **Calculate the number of moles of carbon**:
- The molar mass of carbon is approximately 12 g/mol:
\[
\text{Moles of C} = \frac{2.4003 \, \text{g}}{12 \, \text{g/mol}} \approx 0.2000 \, \text{mol}
\]
7. **Calculate the percentage composition of carbon**:
\[
\text{Percentage of C} = \left( \frac{\text{Mass of carbon}}{\text{Total mass}} \right) \times 100 = \left( \frac{2.4003 \, \text{g}}{2.7 \, \text{g}} \right) \times 100 \approx 88.89\%
\]
Thus, the percentage composition of carbon in the compound is approximately **88.89%**.
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.