Home
Class 11
CHEMISTRY
60 g of water gas and 18 g of steam are ...

60 g of water gas and 18 g of steam are heated in a closed vessel to a temperature of `450^@C` so that the equilibrium : `CO_(g) + H_2O_(g)Leftrightarrow CO_(2(g)) + H_(2(g))` is reached. If K for the reaction is 3, the mass of `CO_2` present will be

A

11g

B

22g

C

44g

D

60g

Text Solution

Verified by Experts

The correct Answer is:
B
Moles of water gas `(CO+H_(2))=(60)/(30)=2`
Moles of CO=Moles of `H_2=1` Moles of steam `=(18)/(18)=1`
`"Initial conc. "underset(1)(CO)+underset(1)(H_(2)O) Leftrightarrow underset(0)(CO_2)+underset(1)(H_(2))`
At equilibrium conc. `"1-x 1-x x 1+x"`
`K=3 ([CO_(2)] [H_(2)])/([CO] [H_(2)O])=(x xx (1+x))/(1-x) (1-x)`
`K=3 ([CO_(2)] [H_(2)])/([CO] [H_(2)O])=(x xx (1+x))/((1-x) (1-x))`
x=0.5 mol of `CO_2=0.5`
Hence, the mass of `CO_2=22g`
Promotional Banner

Similar Questions

Explore conceptually related problems

The equilibrium constant for the reaction is H_(2)O_((l))+CO_((g)) Leftrightarrow H_(2(g))+CO_(2(g)) is 64. If the rate constant for the forward reaction is 160, the rate constant for the backward reaction is

Predict the effect of compression on the following equilibrium reaction: CH_4(g) + H_2O(g) hArr CO(g) +3H_2(g)

For the reaction, CO(g) + H_(2)O(g) rarr CO_(2)(g) + H_(2)(g) , at a given temperature, the equilibrium amount of CO_(2) (g) can be increased by:

For the reaction, CO_((g))+H_(2)O_((g))hArrCO_(2(g))+H_(2(g)) at a given temperature, the equilibrium amount of CO_(2(g)) can be increased by:

Consider the following reaction : CH_(4)(g)+H_(2)O(g) overset(1270 K) to CO(g)+3H_(2)(g) In the reaction given above, the mixture of CO and H_(2) is :