Question

what volume of NH3 (at STP) con be recovered by treating 30 ml of a
O.2N (NH4) SO4 with 4g NaOH ?​

Answer 1

given, 30ml of 0.2N (NH4)2SO4

molarity = normality/n-factor

so, molarity of (NH4)2SO4 = 0.2/2 = 0.1M

now no of moles of (NH4)2SO4 = 0.1 × 30/1000 = 0.1 × 0.03 = 0.003 mol

again, 4g of NaOH = 4/40 mol of NaOH = 0.1 mol of NaOH

reaction between (NH4)2SO4 and NaOH is ...

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

here it is clear that one mole of (NH4)2SO4 reacts with 2 mole of NaOH

so, 0.003 mole of (NH4)2SO4 reacts with 0.006 mole of NaOH.

hence, (NH4)2SO4 is limiting reagent

now, one mole of (NH4)2SO4 produces 2 moles of NH3.

so, 0.003 mole of (NH4)2SO4 produces 0.006 mole of NH3.

so, volume of NH3 = 0.006 mole × volume of one mole of gas at STP

= 0.006 × 22.4 L

= 0.1344L

= 134.4ml