Question

un 15 Complee! If 40.1 g of sulfurous acid reacts with 23.2 g of sodium hydroxide, how many grams of H₂O can be produced? what is the mass of each reactant left after the reaction is over?

Answer 1

Molar mass of  H₂SO₃ = ( 2 $\times$ 1.0079) + 32.06 + ( 3 $\times$ 16.00 ) = 82.08 g/mol

Molar mass of NaOH = 22.99 + 16.00 + 1.0079 = 40.00 g / mol

Molar Mass of H₂O = ( 2 $\times$ 1.0079) + 16.00 = 18.02 g /mol

Consider a balanced chemical equation for given reaction.

H₂SO₃ + 2 NaOH $\rightarrow$ Na₂SO₃ + 2 H₂O

From reaction, 1 mol H₂SO₃$\equiv$ 2 mol NaOH $\equiv$ 1 mol  Na₂SO₃$\equiv$ 2 mol H₂O

Consider relation, 1 mol H₂SO₃$\equiv$ 2 mol NaOH

82.08 g H₂SO₃$\equiv$ 2 $\times$ 40.00 g NaOH

$\therefore$ 40.1 g H₂SO₃$\equiv$ (2 $\times$ 40.00 $\times$ 40.1 / 82.08 ) g NaOH

40.1 g H₂SO₃$\equiv$ 39.08 g NaOH

i e 40.1 g H₂SO₃ will react with 39.08 g NaOH. But the amount of NaOH is 23.2 g . This is less than required amount ( 39.08 g) . Hence, NaOH is limiting reactant. Therefore, yield of products will depend on mass of limiting reactant NaOH.

Consider relation, 2 mol NaOH $\equiv$ 2 mol H₂O

2 $\times$ 40.00 g NaOH $\equiv$ 2 $\times$ 18.02 g H₂O

$\therefore$ 23.2 g NaOH $\equiv$ 2 $\times$ 18.02 $\times$ 23.2 / ( 2 $\times$ 40 ) g H₂O

23.2 g NaOH $\equiv$ 10.45 g H₂O

ANSWER : Mass of water produced in the reaction = 10.45 g

We have , 2 mol NaOH $\equiv$ 1 mol H₂SO₃

2 $\times$ 40.00 g NaOH $\equiv$ 82.08 g H₂SO₃

$\therefore$ 23.2 g NaOH $\equiv$ 82.08 $\times$ 23.2 / 2 $\times$ 40.00 g H₂SO₃

$\therefore$ 23.2 g NaOH $\equiv$ 23.80 g H₂SO₃

Hence, mass of H₂SO₃ remained non reacted = 40.1 g - 23.80 g = 16.3 g

NaOH is limiting reactant , hence it is all consumed in the reaction. Hence, mass of NaOH left after the reaction = 0 g