The mixture of CO and CO2 contains 60% CO and rest is CO2. Two liters of this mixture is passed over red hot charcoal. What is the volume of mixture after reaction

Question

The mixture of CO and CO2 contains 60% CO and rest is CO2. Two liters of this mixture is passed over red hot charcoal. What is the volume of mixture after reaction (A) 3.6 L (B) 2.4 L (C) 2.8 L (D) 4 L

Tardigrade Admin · Accepted Answer

Volume of CO = (60/100) ×2 = 1.2L Volume of CO2 = (40/100) × 2 = 0.8L CO2 + 2C → 2CO 0.8 L CO2 produces = 1.6 L of Co Volume of mixture after the reaction = 1.6 + 1.2 = 2.8 L

Q. The mixture of $CO$ and $C O_{2}$ contains $60$ and rest is $C O_{2}$ . Two liters of this mixture is passed over red hot charcoal. What is the volume of mixture after reaction

$3.6 L$

$2.4 L$

$2.8 L$

$4 L$

Volume of $CO = \frac{60}{100} \times 2 = 1.2 L$
Volume of $C O_{2} = \frac{40}{100} \times 2 = 0.8 L$
$C O_{2} + 2 C \to 2 CO$
$0.8 L C O_{2}$ produces $= 1.6 L$ of $C o$
Volume of mixture after the reaction
$= 1.6 + 1.2 = 2.8 L$

Q. The mixture of CO and CO2​ contains 60 and rest is CO2​. Two liters of this mixture is passed over red hot charcoal. What is the volume of mixture after reaction

3.6L

2.4L

2.8L

4L