The equilibrium constant Kc for the equation2H2(g) + CO(g) ⇌ CH3OH(g)is 35 at a certain temperature. If there are 3.21 ×10−2 moles of H2 and 4.87 ×10−3 moles of CH3OH at equilibrium in a 3.63−L flask, what is the concentration of CO?

the concentration of [CO]= 0.0532M

Explanation:

From The equation of reaction

2H2(g) + CO(g) ⇌ CH3OH(g)

Applying Kc= [CH3OH]/[H2]^2[[CO]

[CH3OH]= 0.00487

[CO]= x-0.00487

[H2]=(0.032-0.00487)^2=0.0271

Substitute into formula

Kc=[CH3OH]/[H2]^2[[CO]

35= 0.00487/(0.0271)^2(x-0.00487)

Simplify

x-0.00487=0.189

x= 0.00487+0.189=0.193moles

[CO]= n/C= 0.193/3.63= 0.0532M

the correct name for the n3 ion is azide. azide is a polyatomic anion with a -1 charge and is written as n3 -1. it consists of three nitrogen atoms bonded together. hope i . let me know if you need anything else.