To determine how many grams of XeF6 are required to react with 0.321 L of hydrogen gas at 4.33 atm and 75°C in the reaction shown below, we can use the ideal gas law, which states that the pressure, volume, and temperature of a gas are related by the equation PV = nRT, where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature of the gas in Kelvin.
In this problem, we know the volume of the hydrogen gas, the pressure of the hydrogen gas, and the temperature of the hydrogen gas, so we can use these values to calculate the number of moles of hydrogen gas present in the reaction:
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n = (PV) / (RT)
n = (4.33 atm * 0.321 L) / (0.08206 L*atm / mol*K * (75 + 273.15) K)
n = 0.025 mol
Next, we need to determine the number of moles of XeF6 required to react with 0.025 mol of hydrogen gas in the reaction shown below. We can do this by using the balanced chemical equation for the reaction, which is XeF6(s) + 3 H₂(g) → Xe(g) + 6 HF(g). From this equation, we can see that for every 1 mole of XeF6 that reacts, 3 moles of hydrogen gas are consumed. Therefore, to consume 0.025 mol of hydrogen gas in the reaction, we need 0.025 mol / 3 = 0.0083 mol of XeF6.
Finally, we need to convert the number of moles of XeF6 to grams. To do this, we can use the molar mass of XeF6, which is 221.0 g/mol. Therefore, 0.0083 mol of XeF6 has a mass of 0.0083 mol * 221.0 g/mol = 1.83 g.
In summary, to react with 0.321 L of hydrogen gas at 4.33 atm and 75°C in the reaction shown below, we need 1.83 g of XeF6.