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