Use Fe = 56, C = 12, O = 16 and molar volume 22.4 L/mol at STP. Excess acidified KMnO4 oxidises 3.6 g of FeC2O4 completely (Fe^2+ -> Fe^3+, C2O4^2- -> 2CO2). Find the volume of CO2 evolved at STP in mL (nearest integer).

JEE Main levelNumerical answerDifficulty 2 of 3rx135

Use Fe = 56, C = 12, O = 16 and molar volume $22.4 L/mol$ at STP. Excess acidified $KMnO X_{4}$ oxidises $3.6 g$ of $FeC X_{2} O X_{4}$ completely ( $Fe X^{2 +} Fe X^{3 +}$ , $C X_{2} O X_{4} X^{2 -} 2 CO X_{2}$ ). Find the volume of $CO X_{2}$ evolved at STP in mL (nearest integer).

Numerical answer type, answer in mL, tolerance 1.

M (FeC X_{2} O X_{4}) = 144

, so moles

= 3.6/144 = 0.025

. Each formula unit gives 2

CO X_{2}

, so

n (CO X_{2}) = 0.05

mol. Volume

= 0.05 \times 22.4 = 1.12 L = 1120 mL