Question

When 6.3 g of sodium bicarbonate is added to 30 g of acetic acid solution, the residual solution is found to weigh 33 g. The number of moles of carbon dioxide released in the reaction is
(a) 3
(b) 0.75
(c) 0.075
(d) 0.3

Answer 1

NaHCO₃ + CH₃COOH ⇒CH₃COONa + CO₂ + H₂O
molecular weight of NaHCO₃ = 84g/mol
Mol. Weight of CH₃COOH = 60g/mol
Mol.weigt of CO₂ = 44g/mol

Here given, weight of NaHCO₃ = 6.3g
weight of CH₃COOH = 30g
according to reaction, we see 84g of NaHCO₃ reacts with 60g of CH₃COOH
but here given Weight of NaHCO₃ < given weight of CH₃COOH
so, NaHCO₃ is limiting reagent.

∵ 84g of NaHCO₃ forms , CO₂ = 44g
∴6.3g of NaHCO₃ forms , CO₂ = 44/84 × 6.3 = 3.3g

so, the number of moles of CO₂ = Given weight/mol.weight
= 3.3/44
= 0.3/4 = 0.075

Hence option (c) is correct

Answer 2

Reactant=36.3g

Product. =33g

Number of g of carbon dioxide

released=36.3-33=3.3g

Molecular mass of carbon dioxide

is =1*12+2*16=44

Released mass/

Molecular mass=3.3/44=0.075