Question

3. The equation for the decomposition of sodium hydrogencarbonate is :

Answer 1

a. 13.25 g

b. 3 L

Further explanation

Given

21 g of Sodium hydrogencarbonate

Reaction

2 NaHCO3 (s) → Na2CO3 (s) + H2O(l) + CO2(g)

Required

mass of residue

volume of CO2

Solution

mol  NaHCO₃ (MW=84 g/mol) :

$\tt mol=\dfrac{21}{84}=0.25$

a. the residue = Na₂CO₃

From the equation, mol ratio NaHCO₃ : Na₂CO₃ = 2 : 1, so mol Na₂CO₃ :

$\tt \dfrac{1}{2}\times 0.25=0.125$

mass Na₂CO₃ (MW=106 g/mol) :

$\tt mass=0.125\times 106=13.25~g$

b. From equation, mol CO₂ = 0.5 x mol NaHCO₃ = 0.125

RTP= 25 C, 1 atm ⇒ 1 mol =24 L

So volume CO₂ :

$\tt 0.125\times 24=3~L$