Suppose that the concentration of ammonia produced in the reaction is 0.7 m and the concentrations of nitrogen and hydrogen used are 0.8 m and 0.02 m respectively. calculate the equilibrium constant for the reaction. start with the expression that follows this equation:
n2 (g) + 3h2 (> 2nh3 (g)
a. 0
b. 1 x 10^-6
c. 100,000
d. 1,000,000

your picture is not part of the scientific method.

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2