Final answer:
To generate 153.0 l of gas, composed of 147.0 l of N₂ and 6.0 l of O₂ at 0.720 atm and 297 K, you would need 28.6 kg of potassium nitrate.
Step-by-step explanation:
To calculate the mass of potassium nitrate needed, we can use stoichiometry and the ideal gas law. First, we need to convert the volumes of N₂ and O₂ to moles using the ideal gas law:
N₂: (147.0 L)(0.720 atm) / (0.0821 L·atm/(mol·K) * 297 K) = 7.08 mol
O₂: (6.0 L)(0.720 atm) / (0.0821 L·atm/(mol·K) * 297 K) = 0.27 mol
Next, we can use the balanced chemical equation and the stoichiometry to find the moles of potassium nitrate needed:
N₂ + O₂ → 2NO
mole ratio of N₂ to NO is 1:2, so 7.08 mol N₂ will produce 14.16 mol NO
So, 14.16 mol NO requires 14.16/0.05 = 283.2 mol of KNO₃.
Finally, we can convert moles to mass:
Mass of potassium nitrate = (283.2 mol)(101.1 g/mol) = 28605 g or 28.6 kg