Question #35382
a)How many grams of NH3 can be produced from 4.12mol of N2 and excess H2.
b)How many grams of H2 are needed to produce 10.71g of NH3?
c)How many molecules (not moles) of NH3 are produced from 7.31×10−4g of H2?
b)How many grams of H2 are needed to produce 10.71g of NH3?
c)How many molecules (not moles) of NH3 are produced from 7.31×10−4g of H2?
Expert's answer
N2+3H2=2NH3
a) n(NH3)=4.12*2=8.24mol; m(NH3)=8.24*17=140.08g
b) n(NH3)=10.71/17=0.63mol; m(H2)=(0.63*3/2)*2=1.89g
c) n(H2)=7.31*10^-4/2=3.655*10^-4mol; n(NH3)=3.655*10^-4*2/3=2.436*10^-4mol;
N=n*Na=2.436*10^-4*6.02*10^23=1.466*10^20 (particles)
a) n(NH3)=4.12*2=8.24mol; m(NH3)=8.24*17=140.08g
b) n(NH3)=10.71/17=0.63mol; m(H2)=(0.63*3/2)*2=1.89g
c) n(H2)=7.31*10^-4/2=3.655*10^-4mol; n(NH3)=3.655*10^-4*2/3=2.436*10^-4mol;
N=n*Na=2.436*10^-4*6.02*10^23=1.466*10^20 (particles)
