The volume in L H₂(g)(measured at 22 °C and 745 mmHg)required to react with 30.0 L CO(g)(measured at 0 °C and 760 mmHg)in the reaction 3 CO(g)+ 7 H₂(g)→ C₃H₈(g)+ 3 H₂O(l)is: A)63.5 L B)74.1 L C)77.2 L D)70.0 L E)33.1 L

Question

Quizplus · Accepted Answer

To solve this question, we need to use the ideal gas law equation: PV = nRT. First, we need to convert the given volumes of CO and H₂ to moles using the ideal gas law equation. For CO: P = 760 mmHg (converted to atm) V = 30.0 L T = 0 + 273 = 273 K Using PV = nRT, we can solve for n: n = PV/RT = (760/760) * (30.0/22.414) * (0.08206 * 273) = 38.2 mol For H₂: P = 745 mmHg (converted to atm) V = ? (what we need to find) T = 22 + 273 = 295 K Using PV = nRT and the stoichiometry of the balanced equation, we can solve for V: 3 mol CO : 7 mol H₂ 38.2 mol CO : n mol H₂ n = (7/3) * 38.2 = 88.9 mol V = nRT/P = (88.9 * 0.08206 * 295)/0.745 = 77.2 L Therefore, the answer is choice C: 77.2 L.

The Volume in L H2(g)(measured at 22 °C and 745

The Volume in L H₂(g)(measured at 22 °C and 745