The volume of CO2 (in cm3) liberated at S.T.P. when 1.06 g of anhydrous sodium carbonate is treated with excess of dilute HCl is [atomic mass of Na= 23]

Question

The volume of CO2 (in cm3) liberated at S.T.P. when 1.06 g of anhydrous sodium carbonate is treated with excess of dilute HCl is [atomic mass of Na= 23] (A) 112 (B) 224 (C) 56 (D) 2240

Tardigrade Admin · Accepted Answer

Na 2 CO 3+2 HCl longrightarrow 2 NaCl + CO 2+ H 2 O 106 g 22400 cm 3 1.06 g (22400/106) × 1.06 cm 2 =224 cm 3

Q. The volume of $C O_{2}$ (in $c m^{3}$ ) liberated at S.T.P. when $1.06 g$ of anhydrous sodium carbonate is treated with excess of dilute $H Cl$ is [atomic mass of $N a = 23$ ]

112

224

56

2240

$N a_{2} C O_{3} + 2 H Cl ⟶ 2 N a Cl + C O_{2} + H_{2} O$
$106 g 22400 c m^{3}$
$1.06 g \frac{22400}{106} \times 1.06 c m^{2}$
$= 224 c m^{3}$

Q. The volume of CO2​ (in cm3) liberated at S.T.P. when 1.06g of anhydrous sodium carbonate is treated with excess of dilute HCl is [atomic mass of Na=23]

112

224

56

2240

Na2​CO3​+2HCl⟶2NaCl+CO2​+H2​O 106g22400cm3 1.06g10622400​×1.06cm2 =224cm3

Q. The volume of $C O_{2}$ (in $c m^{3}$ ) liberated at S.T.P. when $1.06 g$ of anhydrous sodium carbonate is treated with excess of dilute $H Cl$ is [atomic mass of $N a = 23$ ]

$N a_{2} C O_{3} + 2 H Cl ⟶ 2 N a Cl + C O_{2} + H_{2} O$
$106 g 22400 c m^{3}$
$1.06 g \frac{22400}{106} \times 1.06 c m^{2}$
$= 224 c m^{3}$