How many liters of nh 3 are needed to react completely with 30.0 l of no (at stp)?

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

your answer would be “dry”

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