What volume of carbon monoxide,measured at 1200 °C and 720 Torr,would be released when 5.0 kg of an ore containing 45% zinc oxide as the only reacting species is treated with excess carbon? [ZnO(s)+ C(s)→ Zn(s)+ CO(g)] A)1.2 × 10² L B)1.5 × 10³ L C)3.5 × 10³ L D)7.8 × 10³ L E)7.0 × 1 0⁴ L

Question

Quizplus · Accepted Answer

Step 1: Calculate the moles of ZnO in 5.0 kg of the ore.

Mass of ZnO = 0.45 x 5.0 kg = 2.25 kg

Moles of ZnO = (2.25 kg / 81.39 g/mol) = 0.02766 mol

Step 2: From the balanced chemical equation, it is clear that 1 mol of ZnO produces 1 mol of CO. Therefore, the number of moles of CO produced will also be 0.02766 mol.

Step 3: Calculate the volume of CO at the given conditions of temperature and pressure using the ideal gas law:

PV = nRT

V = (nRT) / P

V = [(0.02766 mol) (0.08206 L atm/mol K) (1473 K)] / (720 Torr)

V = 3.47 L or 3.5 x 10^3 mL

Therefore, the volume of CO released will be 3.5 x 10^3 mL or 3.5 L, which corresponds to choice (C).

What Volume of Carbon Monoxide,measured at 1200 °C and 720