Question no. 1.24...plzz solv it..
N2 + 3H2 -------> 2NH3
1 mol + 3 Mol = 2 moles
No. of Moles of N2 = 206/28 = 7.35
No. of moles of H2 = 103/2 = 51.5
From the reaction,
7.35 moles of N2 will react with (7.35x3) = 22.05 M of H2
Thus, H2 is in excess.
a) Mass of N2 + Mass of H2 = 206 g + (22.05 x2) = 206 + 44.1 = 250 g of ammonia
b) H2 remains unreacted
c) Mass of unreacted H2 = (51.5 - 22.05) X 2 = 58.9 g
1 mol + 3 Mol = 2 moles
No. of Moles of N2 = 206/28 = 7.35
No. of moles of H2 = 103/2 = 51.5
From the reaction,
7.35 moles of N2 will react with (7.35x3) = 22.05 M of H2
Thus, H2 is in excess.
a) Mass of N2 + Mass of H2 = 206 g + (22.05 x2) = 206 + 44.1 = 250 g of ammonia
b) H2 remains unreacted
c) Mass of unreacted H2 = (51.5 - 22.05) X 2 = 58.9 g