Lactose (molecular weight 342) is burnt completely according to the reaction: C₁₂H₂₂O₁₁ + 12O₂ → 12CO₂ + 11H₂O. What weight of CO₂ would be formed when 1/12 mole of this compound is burnt?
Correct answer: C. 44 g
- A. 12 g
- B. 4.4 g
- C. 44 g
- D. 440 g
Explanation
According to the balanced equation, 1 mole of lactose produces 12 moles of CO₂. Therefore, 1/12 mole of lactose will produce (1/12) × 12 = 1 mole of CO₂. The mass is 1 mol × 44 g/mol (molar mass of CO₂) = 44 g.
Report an error
The more specific you are, the faster it gets fixed. A source beats an opinion.
Prefer email? [email protected]
About Fundamental Concepts of Chemistry
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.
Practise Fundamental Concepts of Chemistry
894 free Fundamental Concepts of Chemistry MCQs from Chemistry, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Discussion
Stuck on an option, or know a faster way to get there? Ask or explain it here.
No comments yet. Be the first to explain this one.
Exams that ask Chemistry questions like this
Chemistry is on 14 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
More Fundamental Concepts of Chemistry questions
For the reaction A + 2B → C, 5 moles of A and 8 moles of B will produce how many moles of C?
Element A (atomic weight 12.01) and element B (atomic weight 16) combine to form a new substance X. If two moles of B combine with one mole of A, then the weight of one mole of X is:
10 g of carbon burns, giving 11.2 litres of CO₂ at STP. After combustion, the amount of unburnt carbon is:
The amount of a given product calculated to be obtained from a chemical reaction, assuming the reaction goes to completion, is:
A compound (60 g) on analysis gave C = 24 g, H = 4 g, O = 32 g. Its empirical formula is:
An oxide of metal (M) has 40% by mass of oxygen. Metal M has a relative atomic mass of 24. The empirical formula of the oxide is: