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 + C → Zn + CO] 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

First, we need to find the amount of reacting species present in the ore. Since the ore contains 45% zinc oxide, 5.0 kg of ore would contain 2.25 kg of zinc oxide.
Next, we need to find the moles of carbon monoxide produced:
1 mole of ZnO produces 1 mole of CO
2.25 kg of ZnO is equal to 22.32 moles (using molar masses)
Therefore, 22.32 moles of CO are produced.
Now, we can use the ideal gas law to find the volume at 1200°C and 720 torr:
PV = nRT
V = (nRT)/P
V = (22.32 mol x 0.0821 L·atm/mol·K x (1200 + 273) K) / (720 torr x 1 atm/760 torr)
V = 35.2 L
Therefore, the volume of carbon monoxide produced is 3.5 x 10¹² L.

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