Answer:
a) No. of moles of hydrogen needed = 5.4 mol
b) Grams of ammonia produced = 27.2 g
Explanation:
[tex]N_2 (g) + 3H_2(g) \rightarrow 2NH_3 (g)[/tex]
a)
No. of moles of nitrogen = 1.80 mol
1 mole of nitrogen reacts with 3 moles of hydrogen
1.80 moles of nitrogen will react with
= 1.80 × 3 = 5.4 moles of hydrogen
b)
No. of moles of hydrogen = 2.4 mol
It is given that nitrogen is present in sufficient amount.
3 moles of hydrogen produce 2 moles of [tex]NH_3[/tex]
2.4 moles of hydrogen will produce
= [tex]\frac{2}{3} \times 2.4 = 1.6\ mol\ of\ NH_3[/tex]
Molar mass of ammonia = 17 g/mol
Mass in gram = No. of moles × Molar mass
Mass of ammonia in g = 1.6 × 17
= 27.2 g
