Question
How many grams of CaH, are needed to generate 147 L of H2 gas if the pressure of H2 is 826 torr at 20 °C? Express your answer using three significant figures.
Answers
GPT-4o mini
To calculate the grams of calcium hydride (CaH₂) needed to generate 147 L of hydrogen gas (\(H_2\)) at a given pressure and temperature, we can follow these steps:
1. **Use the Ideal Gas Law**: The ideal gas law relates the pressure, volume, temperature, and number of moles of a gas:
\[
PV = nRT
\]
Where:
- \(P\) = pressure (in atm)
- \(V\) = volume (in L)
- \(n\) = number of moles
- \(R\) = ideal gas constant (0.0821 L·atm/(K·mol))
- \(T\) = temperature (in K)
2. **Convert the Given Pressure**:
The given pressure is in torr, so we convert it to atm:
\[
P = \frac{826 \, \text{torr}}{760 \, \text{torr/atm}} \approx 1.088 \, \text{atm}
\]
3. **Convert the Temperature to Kelvin**:
The temperature is given in Celsius, so we convert it to Kelvin:
\[
T(K) = 20 + 273.15 = 293.15 \, K
\]
4. **Calculate the Number of Moles of Hydrogen Gas**:
Rearranging the ideal gas law for \(n\):
\[
n = \frac{PV}{RT}
\]
Substituting the values:
\[
n = \frac{(1.088 \, \text{atm})(147 \, \text{L})}{(0.0821 \, \text{L·atm/(K·mol)})(293.15 \, K)}
\]
\[
n \approx \frac{160.296 \, \text{atm·L}}{24.2481 \, \text{L·atm/(K·mol)}} \approx 6.61 \, \text{mol H}_2
\]
5. **Relate Moles of CaH₂ to Moles of \(H_2\)**:
The reaction for the formation of \(H_2\) from \(CaH_2\) is:
\[
CaH_2 \rightarrow Ca + H_2
\]
This shows that 1 mole of \(CaH_2\) produces 1 mole of \(H_2\). Thus, \(6.61\) moles of \(H_2\) will require \(6.61\) moles of \(CaH_2\).
6. **Calculate the Mass of CaH₂**:
The molar mass of \(CaH_2\) can be calculated as follows:
- Calcium (Ca) = approximately 40.08 g/mol
- Hydrogen (H) = approximately 1.01 g/mol (2 atoms)
\[
\text{Molar mass of } CaH_2 = 40.08 + (2 \times 1.01) = 42.10 \, g/mol
\]
Therefore, the mass of \(CaH_2\) is:
\[
\text{mass} = n \times \text{molar mass} = 6.61 \, \text{mol} \times 42.10 \, g/mol \approx 278.3 \, g
\]
7. **Final Answer**:
Rounding the mass to three significant figures, we find:
\[
\boxed{278 \, \text{g}}
\]
Thus, approximately **278 grams** of CaH₂ are needed to generate 147 L of \(H_2\) gas at the specified conditions.
1. **Use the Ideal Gas Law**: The ideal gas law relates the pressure, volume, temperature, and number of moles of a gas:
\[
PV = nRT
\]
Where:
- \(P\) = pressure (in atm)
- \(V\) = volume (in L)
- \(n\) = number of moles
- \(R\) = ideal gas constant (0.0821 L·atm/(K·mol))
- \(T\) = temperature (in K)
2. **Convert the Given Pressure**:
The given pressure is in torr, so we convert it to atm:
\[
P = \frac{826 \, \text{torr}}{760 \, \text{torr/atm}} \approx 1.088 \, \text{atm}
\]
3. **Convert the Temperature to Kelvin**:
The temperature is given in Celsius, so we convert it to Kelvin:
\[
T(K) = 20 + 273.15 = 293.15 \, K
\]
4. **Calculate the Number of Moles of Hydrogen Gas**:
Rearranging the ideal gas law for \(n\):
\[
n = \frac{PV}{RT}
\]
Substituting the values:
\[
n = \frac{(1.088 \, \text{atm})(147 \, \text{L})}{(0.0821 \, \text{L·atm/(K·mol)})(293.15 \, K)}
\]
\[
n \approx \frac{160.296 \, \text{atm·L}}{24.2481 \, \text{L·atm/(K·mol)}} \approx 6.61 \, \text{mol H}_2
\]
5. **Relate Moles of CaH₂ to Moles of \(H_2\)**:
The reaction for the formation of \(H_2\) from \(CaH_2\) is:
\[
CaH_2 \rightarrow Ca + H_2
\]
This shows that 1 mole of \(CaH_2\) produces 1 mole of \(H_2\). Thus, \(6.61\) moles of \(H_2\) will require \(6.61\) moles of \(CaH_2\).
6. **Calculate the Mass of CaH₂**:
The molar mass of \(CaH_2\) can be calculated as follows:
- Calcium (Ca) = approximately 40.08 g/mol
- Hydrogen (H) = approximately 1.01 g/mol (2 atoms)
\[
\text{Molar mass of } CaH_2 = 40.08 + (2 \times 1.01) = 42.10 \, g/mol
\]
Therefore, the mass of \(CaH_2\) is:
\[
\text{mass} = n \times \text{molar mass} = 6.61 \, \text{mol} \times 42.10 \, g/mol \approx 278.3 \, g
\]
7. **Final Answer**:
Rounding the mass to three significant figures, we find:
\[
\boxed{278 \, \text{g}}
\]
Thus, approximately **278 grams** of CaH₂ are needed to generate 147 L of \(H_2\) gas at the specified conditions.