can some one please help me please fast fast
Chemistry

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N_{2}(g) + 3H_{2}(g) 2NH_{3}(g)
is found to contain 5.27 g of N_{2}, 0.339 g of H_{2}, and 0.0477 g of NH_{3}. Calculate the equilibrium constant (K_{eq} expressed in terms of the molar concentrations) at the given temperature.
2,A 3.00L vessel contained 9.10×10^{2} mol of gaseous PCl_{3}, 9.48×10^{2} mol of gaseous PCl_{5}, and 8.80×10^{2} mol of Cl_{2} gas at equilibrium at 242°C.
Calculate the value of K_{eq} for the reactionPCl_{3}(g) + Cl_{2}(g) PCl_{5}(g)
DATA:
Mass of N_{2}= 5.27g = (0.1882 mol*1000)/1500 =0.1254M
Mass of H_{2}=0.339g =(0.0188 mol*1000)/1500 =0.0125M
Mass of NH_{3}= o.o477g = (0.0028 mol*1000)/1500=0.0018M
K_{eq} =?
SOLUTION:
K_{eq} = [NH3]2/[N_{2}] [H_{2}]^{3}
By putting the molar concentrations of reactants and product in the above equation;
K_{eq} = [.0018]^{2}/[.1254] [.0125]^{3}
= 13.22
gonna send you the next solution.
DATA:
mol of PCl_{3} = (9.10*10^{2} mol*1000)/3000=0.0303M
Mole of PCl_{5} = (9.48*10^{2} mol *1000)/3000=0.0316M
Mole of Cl_{2} = (8.80*10^{2}*1000)/3000= 0.0293M
K_{eq} =?
SOLUTION:
K_{eq} = [PCl_{5}]/[PCl_{3}] [Cl_{2}]
= 0.0316/(0.0303) (0.0293)
K_{eq} = 35.594
DATA:
Mass of N_{2}= 5.27g = (0.1882 mol*1000)/1500 =0.1254M
Mass of H_{2}=0.339g =(0.1695 mol*1000)/1500 =0.113M
Mass of NH_{3}= o.o477g = (0.0028 mol*1000)/1500=0.0018M
K_{eq} =?
SOLUTION:
K_{eq} = [NH3]2/[N_{2}] [H_{2}]^{3}
By putting the molar concentrations of reactants and product in the above equation;
K_{eq} = [.0018]^{2}/[.1254] [0.113]^{3}
= 0.0179
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