High School

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For the following reaction:

\[ \text{C}_3\text{H}_4 + 4\text{O}_2 \rightarrow 3\text{CO}_2 + 2\text{H}_2\text{O} \]

What mass of \(\text{H}_2\text{O}\) will be produced from the combustion of excess hydrocarbon and 35.0 g \(\text{O}_2\)?

A. 19.7 g
B. 9.85 g
C. 39.4 g
D. 31.1 g
E. 62.2 g

Answer :

Final answer:

To find the mass of H2O produced, we need to determine the limiting reagent first. The equation tells us that the stoichiometric ratio of O2 to H2O is 4:2, which simplifies to 2:1. We can use that ratio to convert the mass of O2 given (35.0 g) to the mass of H2O produced. Therefore, the mass of H2O produced is 9.85 g.

Explanation:

To find the mass of H2O produced, we need to determine the limiting reagent first. The equation tells us that the stoichiometric ratio of O2 to H2O is 4:2, which simplifies to 2:1. We can use that ratio to convert the mass of O2 given (35.0 g) to the mass of H2O produced.

Step-by-step:

  1. Calculate the molar mass of O2: 32.0 g/mol
  2. Calculate the moles of O2: 35.0 g / 32.0 g/mol = 1.094 mol
  3. Use the stoichiometric ratio to convert moles of O2 to moles of H2O: 1.094 mol O2 * (2 mol H2O / 4 mol O2) = 0.547 mol H2O
  4. Calculate the molar mass of H2O: 18.02 g/mol
  5. Calculate the mass of H2O produced: 0.547 mol H2O * 18.02 g/mol = 9.85 g

Therefore, the mass of H2O produced is 9.85 g. So, the correct answer is b) 9.85 g.

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