How many liters of O₂ gas at 25°C and 1.00 atm pressure are needed to react with 120.5 grams of potassium hydride according to the chemical equation shown below?
2 KH(s) + O₂(g) → H₂O(g) + K₂O(s)

Question

Quizplus · Accepted Answer

1) Convert the mass of potassium hydride to moles: 120.5 g KH x (1 mol KH/ 39.10 g KH) = 3.08 mol KH 2) Use stoichiometry to find the number of moles of O₂ gas needed: 2 mol KH : 1 mol O₂ gas 3.08 mol KH : x x = 1.54 mol of O₂ gas 3) Use the ideal gas law equation to find the volume of O₂ gas at 25°C and 1.00 atm pressure: PV = nRT V = nRT/P V = (1.54 mol) (0.0821 L*atm/mol*K) (298 K) / (1.00 atm) V = 36.7 L Therefore, the best choice is B) 36.7 L because it correctly calculates the volume of O₂ gas needed to react with 120.5 grams of potassium hydride.

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