Moderate

Lactose commonly used as a binder in tablets has a molecular weight of 342. What weight of CO2 would be formed when 1/12 mole of this compound is burnt completely?(C12H22O11 + 12O2 12CO2 + 11H2O)

Correct answer: C. 44 g

  • A. 12 g
  • B. 4.4 g
  • C. 44 g
  • D. 440 g

Explanation

The balanced equation is: C12H22O11 + 12O2 → 12CO2 + 11H2O From the balanced equation, we can see that the molar ratio between lactose (C12H22O11) and CO2 is 1:12, meaning that 1 mole of lactose reacts to form 12 moles of CO2. First, let's determine the number of moles of lactose: Molar mass of lactose (C12H22O11) = 12(12.01 g/mol) + 22(1.01 g/mol) + 11(16.00 g/mol) = 342.34 g/mol Moles of lactose = 1/12 mole (given) Since the molar ratio between lactose and CO2 is 1:12, we can conclude that 1/12 mole of lactose will produce 12/12 = 1 mole of CO2. Now, let's calculate the weight of CO2: Molar mass of CO2 = 12.01 g/mol (carbon) + 16.00 g/mol (oxygen) + 16.00 g/mol (oxygen) = 44.01 g/mol Weight of CO2 = moles of CO2 × molar mass of CO2 = 1 mole × 44.01 g/mol = 44.01 g Thus, the weight of carbon dioxide (CO2) formed when 1/12 mole of lactose is completely burnt is 44.01 grams. Therefore, the correct answer is 44 g.

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Mole calculations connect mass, number of particles and Avogadro's number, while balanced equations provide the mole ratios used in stoichiometry. Questions cover limiting and excess reactants, theoretical yield and percentage yield, including identifying which reactant is consumed first and comparing the actual product with the maximum possible product.

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