Question

The mass of baco3 produced when excess co2 is bubbled through a solution of 0.1025 l of 2 m ba(oh)2 is: ba(oh)2 + co2 baco3 +h2o (molar mass of ba = 137)

Answer 1

The reaction will be :
Ba(OH)2 + CO2 →BaCO3 + H2O
It can be calculated by using POAC (Principle of atomic conservation) :
Barium atoms in Ba(OH)2 in reactant = Barium atoms in BaCO3 in product
Number of moles of BaCO3 = 0.205 mol
​Molar mass of BaCO3 :
137.4 + 12 + 3×16 = 197.4 g

Hence , mass of BaCO3 = Number of moles ×Molar massMass = 0.205 ×197.4 = 40.467 g

Answer 2

Answer:

answer is 40.385g

Explanation:

Ba(OH)2     +     CO2      ----->    BaCO3    +    H2O

1mole                                              1mole

0.205mole                                    0.205mol

THEREFORE,

Molecular mass of BaCO3    =  137 + 12 + 48

                                               =   197g/mol

Weight of BaCO3    =  197*0.205

                                 =  40.385g