The mass of BaCO3 produced when excess CO2 is bubbled through a solution of 0.205 mol Ba(OH)2 is

Question

The mass of BaCO3 produced when excess CO2 is bubbled through a solution of 0.205 mol Ba(OH)2 is (A) 81 g (B) 40.5 g (C) 20.25 g (D) 162 g

Tardigrade Admin · Accepted Answer

Ba(OH)2 + CO2 → Ba CO3 + H2O Atomic wt. of BaCO3 = 137 + 12 + 16 × 3 = 197 No. of mole =(wt. of substance/mol wt) ∵ 1 mole of Ba(OH)2 gives 1 mole of BaCO3 ∴ 0.205 moles of Ba(OH)2 will give 0.205 mole of BaCO3 ∴ wt. of 0.205 mole of BaCO3 will be 0.205 × 197= 40.385 g≈ 40.5 g

Q. The mass of $B a C O_{3}$ produced when excess $C O_{2}$ is bubbled through a solution of $0.205 m o l B a (O H)_{2}$ is

$81 g$

$40.5 g$

$20.25 g$

$162 g$

Q. The mass of BaCO3​ produced when excess CO2​ is bubbled through a solution of 0.205molBa(OH)2​ is

81g

40.5g

20.25g

162g

Ba(OH)2​+CO2​→BaCO3​+H2​O Atomic wt.ofBaCO3​=137+12+16×3=197 No.ofmole=molwtwt.ofsubstance​ ∵1mole of Ba(OH)2​ gives 1 mole of BaCO3​ ∴0.205molesofBa(OH)2​ will give 0.205moleofBaCO3​ ∴wt.of0.205 mole of BaCO3​ will be 0.205×197=40.385g≈40.5g

Q. The mass of $B a C O_{3}$ produced when excess $C O_{2}$ is bubbled through a solution of $0.205 m o l B a (O H)_{2}$ is

$81 g$

$40.5 g$

$20.25 g$

$162 g$