How many mL of 0.122M MnO4- are required?

2016-11-22 12:09 pm

How many mL of 0.122M MnO4- are required to react with an excess of C2O4^2- in order to produce 75.25mL of carbon dioxide gas measured at 30.0C and 715 torr?

MnO4- + C2O4^2- yields MnO2 + CO2 (basic)

回答 (1)

胡雪8°

2016-11-22 1:50 pm

✔ 最佳答案

For the CO₂ gas formed :
Pressure, P = (715/760) atm
Volume, V = 75.25 mL = 0.07525 L
Absolute temperature, T = (273.2 + 30.0) K = 303.2 K
Gas constant, R = 0.08206 L atm / (mol K)
No. of moles, n = ? mol

PV = nRT
Then, n = PV/(RT)
No. of moles of CO₂, n = (715/760) × 0.07525 / (0.08206 × 303.2) mol

2MnO₄⁻ + 3C₂O₄²⁻ + 8H⁺ → 2MnO₂ + 4H₂O + 6CO₂
Mole ratio MnO₄⁻ : CO₂ = 2 : 6 = 1 : 3

No. of moles of MnO₄⁻ = [(715/760) × 0.07525 / (0.08206 × 303.2)] × (1/3) mol

Volume of MnO₄⁻
= {[(715/760) × 0.07525 / (0.08206 × 303.2)] × (1/3) mol} / (0.122 mol/L)
= 0.00777 L
= 7.77 mL

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