When ${CO}_{2} (g)$ is passed over red hot coke it partially gets reduced to $CO (g) . Upon passing 0 .5 L of {CO}_{2} (g)$ over red hot coke, the total volume of the gases increased to 700 mL. The composition of the gaseous mixture at STP is

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CO₂,=300mL;CO=400mL

CO₂,=0.0mL;CO=700 mL

CO₂,=200mL;CO=500mL

CO₂,=350mL;CO=350mL

answer is A.

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Detailed Solution

${CO}_{2} + C ⟶ 2 CO Stoichoimetry ratio is 1 : 2 AT STP, P = 1 atm, T = 273 K, R = 0.0821 Initial moles of {CO}_{2}; n ({CO}_{2} initial) = \frac{PV}{RT} = \frac{1 \times 0 .5}{0 .0821 \times 273} = 0 .022 mole In final mixture no. of moles; n ({CO}_{2} / CO mixture) = \frac{1 \times 0 .7}{0 .0821 \times 273} = 0 .031 Increase in volume is by = 0 .031 - 0 .022 = 0 .009 mole of gas Final no. of moles of CO i.e. n_{(CO final)} n_{(CO final)} = 2 n_{({CO}_{2} initial)} - n_{({CO}_{2} final)} = 2 (0 .022 - n_{({CO}_{2} final)}) . . (i) n_{(CO final)} = 0 .044 - 2 n_{({CO}_{2} final)} . . (ii) ∴ Now, n_{(CO final)} + n_{(CO final)} = 0 .031 n_{({CO}_{2} final)} = 0 .031 - n_{(CO final)} \dots ((ii) Substituting (ii) in eq. (i) n_{(CO final)} = 0 .044 - 2 [0 .031 - n_{(CO final)]} n_{(CO final)} = 0 .044 - 0 .062 + 2 n_{(CO final)} n_{(CO final)} = 0 .018 mol Volume of CO = V = \frac{nRT}{P} = \frac{0 .018 \times 0 .0821 \times 273}{1} = 0 .40 Litre and volume of {CO}_{2} = 0 .7 litre - 0 .4 litre = 0 .3 litre ∴ {CO}_{2} = 300 mL, CO = 400 mL$