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Given the balanced chemical equation:

\[ \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} \]

How many grams of water vapor can be produced if you begin with 37.1 grams of each reactant?

Answer :

If, we begin with 37.1 grams of each reactant, approximately 60.64 grams of water vapor can be produced.

To determine the grams of water vapor produced in the given reaction, we need to use the stoichiometry of the balanced equation. The balanced equation is;

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

According to the balanced equation, for every 1 mole of C₃H₈ reacted, we obtain 4 moles of H₂O produced.

First, let's calculate the number of moles of C₃H₈ in 37.1 grams:

Molar mass of C₃H₈ = 3(12.01 g/mol) + 8(1.01 g/mol) = 44.11 g/mol

Moles of C₃H₈ = 37.1 g / 44.11 g/mol

≈ 0.841 mol

Since the molar ratio between C₃H₈ and H₂O is 1:4, the moles of water vapor produced will be;

Moles of H₂O = 4 moles H₂O/mol C₃H₈ × 0.841 mol C₃H₈

≈ 3.364 mol H₂O

Now, to find the mass of water vapor produced, we need to multiply the moles of H₂O by its molar mass:

Molar mass of H₂O = 2(1.01 g/mol) + 16.00 g/mol

= 18.02 g/mol

Mass of H₂O = Moles of H₂O × Molar mass of H₂O

= 3.364 mol × 18.02 g/mol

≈ 60.64 g

Therefore, if we begin with 37.1 grams of each reactant, approximately 60.64 grams of water vapor can be produced.

To know more about water vapor here

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