Chem question

the synthesis of ammonia is given by the equation:
N2(g) + 3H2(g) <=> 2NH3(g)

at 750K and 900 atm, 3 mol hydrogen and 1 mol nitrogen reacted together to give an equilibrium mixture containing 1.5 mol of ammonia.

a) calculate the number of moles of nitrogen and hydrogen that have reacted

b)hence calculate the number of moles of nitrogen and hydrogen at equilibrium

c) calculate the volume of the equilibrium mixture from the total number of moles present (use R=0.0821 dm^3 atm K^-1 mol^-1)

i would be grateful if someone could help me answer this question or show me where to begin

how did u do that? i dont understand the answer in part A)

You know you have 1.5 moles of ammonia formed. To get the amount of N2 reacted, divide by 2 as the mole ration is 2:1. To get the number of moles of H2, divide by 2 and multiply by 3 as the mole ratio is 3:2. To get the equilibrium values, subtract ur answers in A from the inital number of moles thus.

N2- 0.75-1=0.25
H2- 2.25-3=0.75

thanks very much