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(a) Define the terms:(i) Limiting reagent (ii) Mole(b) 20 g of CaCO3and 20 g of H2SO4 react to give CaSO4 along with water and CO2.(i) Determine the limiting reagent for the above reaction.(ii) How much CaSO4 will be formed?(iii) If 1 mole of gas occupies 22.4 L at STP then calculate the volume of CO2 evolved in the above reaction. [Ca = 40, C = 12, O = 16, H = 1, S = 32] |
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Answer» (a) (i) Limiting reagent: It is the reactant which is completely used up when a reaction goes to completion. (ii) Mole: It is the amount of substance which contains same number of particles (atoms, molecules or ions) as the number of atoms present in 12 g of carbon-12. (b) CaCO3 + H2SO4 → CaSO3 + H2O + CO2 (i) 1 mole CaCO3 requires 1 mole H2SO4 100 g CaCO3 require 98 g H2SO4 \(\therefore\) 20 g CaCO3 will require \(\frac{98}{100}\) x 20 = 19.6 g H2SO4 Hence, CaCO3 is the limiting reagent. (ii) 1 mole, CaCO3 produces 1 mole CaSO4 100 g CaCO3 produce 136 g CaSO4 20 g CaCO3 will produce \(\frac{136}{100}\) x 20 = 27.2 g CaSO4 (iii) 1 mole CaCO3 gives 1 mole CO2 \(\therefore\) 0.2 moles of CaCO3 will give 0.2 moles of CO2 = 0.2 × 22.4 L = 4.48 L of CO2 |
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