Consider the reaction N2(g) + 3 H2(g) <=> 2 NH3(g), which is exothermic as written. What would be the effect on the equilibrium position of adding N2(g)?

Shift to the right side

Explanation:

Remember that any stress that we bring to a chemical equilibrium, will be answered by this in such a way as to minimize the disturbance and attain equilibrium again. This is known as LeChatelier´s principle.

If we add N₂(g) to the system at equilibrium

N₂ ( g) + 3 H₂(g)   ⇄ 2 NH₃(g)

the equilibrium will consume part of the added N₂ gas until it reaches equilibrium again producing more ammonia (NH₃).

Mathematically this is equivalent to a  decrease the pressure of N₂ so that the equilibrium partial pressures of the gases obey the equilbrium pressure constant Kp.

Therefore the equilibrium will shift to the product side.

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