How many milliliters of ammonium sulfate solution having a concentration of 0.218 M are needed to react completely with 62.6 mL of 2.31 M sodium hydroxide solution?

330 mL of (NH₄)₂SO₄ are needed

Explanation:

First of all, we determine the reaction:

(NH₄)₂SO₄  +  2NaOH →  2NH₃  +  2H₂O  +  Na₂SO₄

We determine the moles of base:

(First, we convert the volume from mL to L) → 62.6 mL . 1L/1000 mL = 0.0626L

Molarity . volume (L) = 2.31 mol/L . 0.0626 L = 0.144 moles

Ratio is 2:1. Therefore we make a rule of three:

2 moles of hydroxide react with 1 mol of sulfate

Then, 0.144 moles of NaOH must react with (0.144 .1) /2 = 0.072 moles

If we want to determine the volume → Moles / Molarity

0.072 mol / 0.218 mol/L = 0.330 L

We convert from L to mL → 0.330L . 1000 mL/1L = 330 mL

ill try to . no promises tho.

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