Question

For the following reaction, 5.38 grams of nitrogen gas are allowed to react with 4.79 grams of hydrogen gas . nitrogen(g) + hydrogen(g) ammonia(g) What is the maximum mass of ammonia that can be formed? grams What is the FORMULA for the limiting reagent? What mass of the excess reagent remains after the reaction is complete? grams

Answer 1

Answer – Given, mass of N₂ = 5.38 g , mass of H₂ = 4.79 g

Reaction – N₂ + 3 H₂ ----> 2NH₃

First we need to calculate moles of each –

Mole of N₂ = 5.38 g / 28.014 g.mol^-1

                   = 0.192 moles

Mole of H₂ = 4.79 g / 2.016 g.mol^-1

                   = 2.38 moles

Calculation of limiting reactant –

Moles of NH₃ from N₂ –

From the above balanced equation –

1 moles of N₂ = 2 moles of NH_­3

0.192 moles of N₂ = ? moles of NH_­3

= 0.384 moles of NH_­3

Moles of NH₃ from H₂ –

From the above balanced equation –

3 moles of H₂ = 2 moles of NH_­3

2.38 moles of H₂ = ? moles of NH_­3

= 1.58 moles of NH_­3

So moles of NH₃ gets lowest from the N₂, so limiting reactant is N₂ and moles of NH₃ is considered from the moles of N₂

So, moles of NH₃ = 0.384 moles

Mass of NH₃ = 0.384 moles * 17.0307 g/mol

                      = 6.54 g of NH₃ formed

Mass of excess reactant-

1 moles of N₂ = 3 moles of H₂

So, 0.192 moles of N₂ = ? moles of H₂

= 0.576 moles of H_­2 used in the reaction

So remaining moles of H₂ = 2.38 moles – 0.576 moles

                                           = 1.80 moles

So, mass of excess reactant, H₂ = 1.80 moles * 2.016 g/mole

                                                    = 3.63 g of excess reactant

So, 6.54 g ammonia that can be formed. The formula for the limiting reagent is N₂ .The mass of the excess reagent remains after the reaction is complete is 3.63 g