Answer:
20.8 Lof NH₃ are needed for the reaction
Explanation:
This is the reaction:
4 NH₃ + 6 NO → 5N₂ + 6H₂O
and the info. we need
Ammonia density = 0,00073 g/mL
Let's determine the moles of NO at STP by the Ideal Gases Law equation
P . V = n . R .T
1atm . 30L = n . 0.082 . 273K
(1atm . 30L) / (0.082 . 273K) = n → 1.34 moles of NO
Let's find out the amount of ammonia that should react
6 mol of NO react with 4 mol of ammonia
1.34 mol of NO will react with (1.34 .4)/6 = 0.893 moles of ammonia
Molar mass NH₃ = 17 g/m
0.893 mol . 17 g/m = 15.19 g of ammonia
Ammonia density = 0,00073 g/mL = NH₃ mass / NH₃ volume
0,00073 g/mL = 15.19 g / NH₃ volume
NH₃ volume = 15.19 g / 0,00073 g/mL → 20805.5 mL ⇒ 20.8 L
