Answer:
5.603 g H2
Step-by-step explanation:
M(NH3) = 14 + 1*3 = 17 g/mol
M(H2) = 2*1 = 2 g/mol
3H2 + N2 -----> 2NH3
from reaction 3 mol 2 mol
from reaction 3*2 g 2*17 g
given x g 31.75 g
We can write a proportion.
(3*2)/x = 2*17/31.75
(2*17)x = (3*2)*31.75
x = (3*2)*31.75/(2*17) = 3*31.75/17 =5.603 g H2
