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Some Basic Concepts of Chemistry · Question 21

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1

Butane reacts with oxygen to produce carbon dioxide and water following the equation given below. C4H10(g) + 132O2(g) 4CO2(g) + 5H2O(l) If 174.0 kg of butane is mixed with 320.0 kg of O2, the volume of water formed in liters is . (Nearest integer) [Given : (a) Molar mass of C, H, O are 12, 1, 16 g mol-1 respectively, (b) Density of water = 1 g mL-1]

Answer

138

Solution

Molar mass of C4H10 = 58 g mol-1, of O2 = 32 g mol-1

Moles of C4H10 = 174.0 10358 = 3000 mol

Moles of O2 = 320.0 10332 = 10000 mol

O2 needed for 3000 mol butane = 3000 132 = 19500 mol > 10000 mol

O2 is the limiting reagent.

Moles of H2O = 10000 513/2 = 10000013 = 7692.3 mol

Mass of H2O = 7692.3 18 = 1.3846 105 g

Volume = 1.3846 105 mL = 138.46 L 138 L

Answer: 138

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  1. Pr
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  3. Po
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  1. 0.24 mg
  2. 235.7 g
  3. 2.444 g
  4. 236 mg
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6

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CaCO3(s)+2HCl(aq) CaCl2(aq)+CO2(g)+H2O(l) Consider the above reaction, what mass of CaCl2 will be formed if 250 mL of 0.76 M HCl reacts with 1000 g of CaCO3 ? (Given : Molar mass of Ca, C, O, H and Cl are 40, 12, 16, 1 and 35.5 g mol-1, respectively)

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  3. 10.545 g
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  1. 60
  2. 45
  3. 30
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  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
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