Question

Given the following chemical reactio equation

2NH₃(aq)+H₂SO₄(aq)-->(NH₄)₂SO₄(aq)

Calculate the grams of excess reactant when 3.0g of NH₃ (17.034amu) reacts with 5.0g of H₂SO₄(98.086amu)? Please show work

Answer 1

2NH₃(aq)+H₂SO₄(aq)-->(NH₄)₂SO₄(aq)

Number of moles of NH3 = mass of NH3/molar mass of NH3 = 3/17.034 = 0.17611 moles

Number of moles of H2SO4 = mass of H2SO4/molar mass of H2SO4 = 5.0/98.086 = 0.05097 moles

1 mole of H2SO4 reacts with 2 moles of NH3

Hence 0.05097 moles of H2SO4 will require 0.05097 * 2 = 0.10195 moles of NH3

Moles of NH3 left = 0.17611 - 0.10195 = 0.074158 moles

Mass of NH3 left = number of moles * molar mass

=> 0.074158 moles * 17.034 gm/mol

=> 1.2632 gms