Question

How many grams of NH3

form when 22.3 L
of H2(g)
(measured at STP) reacts with N2
to form NH3
according to this reaction?

N2(g)+3H2(g)→2NH3(g)

Answer 1

Approximately 16.96 grams of NH3 would form when 22.3 liters of H2 reacts with N2 according to the given reaction.

Step-by-step explanation:

In order to determine the number of grams of NH3 formed, we first need to calculate the number of moles of NH3 produced using the given volume of H2. According to the balanced equation, 3 moles of H2 react to form 2 moles of NH3. Therefore, we can use the molar ratio to find the moles of NH3 produced: (22.3 L H2)(1 mol H2/22.4 L H2)(2 mol NH3/3 mol H2) = 0.997 mol NH3.

Next, we can use the molar mass of NH3 (17.03 g/mol) to convert moles of NH3 to grams: (0.997 mol NH3)(17.03 g NH3/1 mol NH3) = 16.96 g of NH3.

So, approximately 16.96 grams of NH3 would form when 22.3 liters of H2 reacts with N2 according to the given reaction.

Answer 2

11.2823 grams of NH3 are produced
Step-by-step explanation:

No. of moles for H2 = 22.3/22.4 = 0.9955 moles

By calculating number of moles produced of NH3, by using ratios:

N2 + 3H2 → 2NH3
1 : 3 : 2
? : 0.9955: ?

part value = 0.9955/3 = 0.33183
No. of moles for NH3 = 2 * 0.33183 = 0.6637 moles

mass of produced NH3 (Molar mass = 17 g/mol) = 0.6637*17 = 11.2823 grams