Question

When ammonia (NH3) reacts with fluorine (F2), the products are dinitrogen tetrafluoride and hydrogen fluoride. 2NH3(g)+5F2(g)→N2F4(g)+6HF(g) How many moles of NH3 are needed to produce 3.00 mol of HF? How many moles of F2 are needed to produce 3.00 mol of HF? How many grams of F2 are required to react with 26.5 g of NH3? How many grams of N2F4 can be produced when 3.55 g of NH3 reacts?

Answer 1

1.) 2 moles or NH3 gives 6 moles of HF.

1 mole of NH3 will give 6/2 moles of HF = 3 moles NH3

Moles of NH3 needed to give 3 moles HF = 1 mole

2.) 5 moles of F2 gives 6 moles HF.

1 mole F2 will give 6/5 moles HF = 1.2 moles.

1 mole HF will be formed by 5/6 moles F2 = 0.833 moles.

Moles of F2 needed to give 3 mol HF = 3 × 0.833. = 2.5 moles F2.

3.) Number of moles = mass/molar mass

Molar mass of NH3 = 17.031 g/mol

Moles of NH3 = 26.5/17.031

= 1.56 moles.

5 moles of F2 reacts with 2 moles of NH3.

1 mole of NH3 will need 2.5 moles F2.

Moles of F2 needed to react 1.56 moles NH3 = 1.56 × 2.5

= 3.9 moles F2.

Molar mass of F2 = 38 g/mol

Mass of F2 needed = 38 g/mol × 3.9 mol

= 148.2 grams F2.

4. 2 moles NH3 gives 1 moles N2F4.

1 mole NH3 will give 0.5 mole N2F4

Molar mass of NH3 = 17.031 g/mol

Moles of NH3 = 3.55/17.031

= 0.208 moles NH3.

Moles of N2F4 formed = 0.208 × 0.5

= 0.104 moles N2F4.

Molar mass of N2F4 = 104.01 g/mol

Mass of N2F4 formed = 104.01 g/mol × 0.104 mol

= 10.82 grams N2F4