The decomposition of a certain mass of CaCO3 gave 11.2 dm3 of CO2 gas at STP. The mass of KOH required to completely neutralise the gas is
(A) 56 g
(B) 28 g
(C) 42 g
(D) 20 g
saurabh kumar , 11 Years ago
Grade 12
2 Answers
R Shyam Sundar
Last Activity: 5 Years ago
For complete neutralisation,
2KOH+CO2 gives K2CO3 + H2O
For 11.2dm^3 i.e.,11.2L(half mole) we need 1mole KOH.
So weight of KOH is 56g
Vikas TU
Last Activity: 5 Years ago
Dear student Weight of 11.2 dm^3 of CO2 gas at STP = 44/2 = 22 gram KOH + CO2 -> KHCO3
KOH required = 56/44 * 22 = 28 gram
B is correct ans. Hope this helps Good Luck
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