Chemistry · Chemical equations and stoichiometry
The decomposition of certain mass of gave of gas at STP. The mass of KOH require
The decomposition of certain mass of \( \mathrm{CaCO}_{3} \) gave \( 11.2 \mathrm{dm}^{3} \) of \( \mathrm{CO}_{2} \) 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
Step-by-step solution
At STP, 1 mole of gas occupies 22.4 dm³. Thus, 11.2 dm³ CO₂ corresponds to 0.5 moles. The neutralization reaction is CO₂ + 2KOH → K₂CO₃ + H₂O, so 1 mole CO₂ requires 2 moles KOH. Therefore, 0.5 moles CO₂ require 1 mole KOH. Molar mass of KOH is 56 g/mol, so mass = 1 × 56 = 56 g.
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