Question

# 90 g of butane on combustion with 80 g of oxygen gave carbon dioxide and water. The gas formed in the reaction is reacted with 40 g of calcium carbonate. Find the amount of product [Ca(HCO3)2] formed. C4H10+O2→CO2+H2OCaCO3+CO2+H2O→Ca(HCO3)2 (Molar mass of Ca = 40 g/mol)

A
44.5 g
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B
162.1 g
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C
64.8 g
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D
120.8 g
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Solution

## The correct option is C 64.8 g2C4H10+13O2→8CO2+10H2O CaCO3+CO2+H2O→Ca(HCO3)2 Moles of C4H10=given massmolar mass=9058=1.55 molMoles of O2=given massmolar mass=8032=2.5 mol Finding the limiting reagent: 2 moles of butane react with 13 moles of oxygen. 1.55 moles will react with 132×1.55=10.075 mol So, oxygen will be the limiting reagent. 13 moles of oxygen produce 8 moles of CO2 2.5 moles will produce = 813×2.5=1.54 moles of CO2 In reaction(ii), Moles of CaCO3=given massmolar mass=40100=0.4 mol CaCO3 will be the limiting reagent. 1 mole of CaCO3 produces 1 mole of Ca(HCO3)2 So, the amount of Ca(HCO3)2 formed = 0.4 mol Mass of Ca(HCO3)2 = 0.4 mol × 162 = 64.8 g

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