Answer:
9.36 grams NaOH needed
Explanation:
Given Rxn
(NH₄)₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2NH₃(g) + 2H₂O shows a molar reaction ratio of 1-mole (NH₄)₂SO₄ to 2-moles NaOH. If the 15.4 gram sample of (NH₄)₂SO₄ is converted to moles, then 2 times moles of ammonium sulfate will be the molar amount of NaOH needed to completely react. Complete computation by converting moles to grams NaOH; that is, multiply by formula wt. of NaOH (=40g/mol).
moles (NH₄)₂SO₄ = 15.4g/132.14g·mol⁻¹ = 0.117 mole (NH₄)₂SO₄.
moles NaOH = 2 x 0.117-mole = 0.234-mole NaOH needed.
∴grams NaOH needed = 0.234-mole NaOH x 40-g NaOH/mole NaOH = 9.36 grams NaOH needed. (3 sig. figs.)
Then multiply 0.117 mole by 2 => 0.234 mole NaOH needed to completely react with the 0.117 mole (NH₄)₂SO₄. Converting to mole to grams, multiply by formula wt. of NaOH => 0.117 mole x 40g/mole = 9.36 grams NaOH needed.
