What volume of CO2 at s.t.p would be obtained by reacting trioxocarbonate (IV) with excess acid?
(G.M.V at s.t.p = 22.4 dm3)
The correct answer is: B
Explanation
Na2CO3 + 2HCC → 2NaII + H2O1000 cm3 IM Na2CO3 contains 22400 cm3
10cm3 IM Na2CO3 will contain (10) / (100) x 224
=224 cm3
10 cm3 0.1 Na2CO3 will contain (o.1 x 224) / (1) = 22.4 cm3