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:
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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)
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Convert the Given Pressure: The given pressure is in torr, so we convert it to atm: \[ P = \frac{826 , \text{torr}}{760 , \text{torr/atm}} \approx 1.088 , \text{atm} \]
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Convert the Temperature to Kelvin: The temperature is given in Celsius, so we convert it to Kelvin: \[ T(K) = 20 + 273.15 = 293.15 , K \]
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Calculate the Number of Moles of Hydrogen Gas: Rearranging the ideal gas law for \(n\): \[ n = \frac{PV}{RT} \] Substituting the values: \[ n = \frac{(1.088 , \text{atm})(147 , \text{L})}{(0.0821 , \text{L·atm/(K·mol)})(293.15 , K)} \] \[ n \approx \frac{160.296 , \text{atm·L}}{24.2481 , \text{L·atm/(K·mol)}} \approx 6.61 , \text{mol H}_2 \]
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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 \(CaH_2\).
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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: \[ \text{mass} = n \times \text{molar mass} = 6.61 , \text{mol} \times 42.10 , g/mol \approx 278.3 , g \]
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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.