Question

calculate the mass of ammonia produced when 8 grams of nitrogen react with 10 grams of hydrogen?which is the reactant that remains after this reaction?by how much gram?

Answer 1

$N_{2} + 3 H_{2} --------> 2N H_{3}$
 number of mole of $N_{2}$ is = 8/28 = 2/7 mole
number of mole of $H_{2}$ is = 10/2 = 5 mole
for 1 mole of $N_{2}$ 3 mole of $H_{2}$ is required
so for 2/7 mole of $N_{2}$ 6/7 mole of $H_{2}$ is required
number of mole of ammonia produce = 2*number of mole of $N_{2}$ consume.
so number of mole of ammonia produce = 2*2/7
mass of ammonia produce = 4*17/7 = 68/7 gm.
moles of $H_{2}$ left = 5 - 3*2/7 = 29/7 mole
mass of $H_{2}$ left = 2*29/7 = 58/7 gm   

Answer 2

MAss of ammonia is 9.7142g
Hydrogen will be remaining (around 8.29g)