at a certain temperature, 0.380 mol ch4 and 0.877 mol h2o is placed in a 1.50 l container.
ch4(g)+2h2o(g)↽−−⇀co2(g)+4h2(g)
at equilibrium, 8.01 g co2 is present. calculate kc.

Kc = 1.4752

Explanation:

⇒ Kc = ( [ CO2(g) ] * [ H2 ]∧4 ) / ( [ CH4 ] * [ H2O ]² )

∴ [ CO2 ] = ( 8.01 g * ( mol / 44.01 g ) ) / 1.5 L

⇒ [ CO2 ] = 0.1213 M

⇒ [ H2O ] = 0.877 mol / 1.5 L = 0.585 M

⇒ [ CH4 ] = 0.380 mol / 1.5 L = 0.253 M

⇒ [ H2 ] = (( 0.380 mol CH4 ) * ( 4 mol H2 / mol CH4 )) / 1.5 L = 1.013 M

⇒ Kc = (( 0.1213 ) * ( 1.013 )∧4 ) / (( 0.253 ) * ( 0.585 )² )

⇒ Kc = 1.4752

it is stable as it is in group 0 which have full outer shells of electrons