Nitrogen and hydrogen gases are combined at high temperatures and pressures to produce ammonia, NH3. If 101.7 g of N2 are reacted with excess H2, how many moles of NH3 will be formed

7.26 moles of NH₃ are formed in this reaction

Explanation:

This is about the reaction for the production of ammonia

1 mol of nitrogen gas reacts to 3 moles of hydrogen in order to produce 2 moles of ammonia.

The equation is: N₂ + 3H₂ → 2NH₃

In the question, we were informed that the excess is the H₂ so the N₂ is limiting reagent. We determine the moles, that has reacted:

101.7 g / 28 g/mol = 3.63 moles

So, If 1 mol of nitrogen gas can produce 2 moles of ammonia

3.63 moles of N₂ must produce ( 2 . 3.63) / 1 = 7.26 moles of NH₃

In this reaction, 7.26 moles of NH3 will be formed.

Explanation:

Step 1: Data given

Mass of N2 = 101.7 grams

Molar mass N2 = 28.0 g/mol

H2 is in excess

Molar mass H2 = 2.02 g/mol

Molar mass of NH3 = 17.03 g/mol

Step 2: The balanced equation

N2(g) + 3H2(g) → 2NH3(g)

Step 3: Calculate moles N2

Moles N2 = mass N2/ molar mass N2

Moles N2 = 101.7 grams / 28.0 g/mol

Moles N2 = 3.63 moles

Step 4: Calculate moles NH3

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

For 3.63 moles N2 we'll produce2*3.63 = 7.26 moles NH3

In this reaction, 7.26 moles of NH3 will be formed.