How many grams of oxygen gas are required to produce 65.75 grams of steam? 2C3H18 (1) + 2502 (g) --> 16CO2 (g) + 18H20 (g) (balanced)

The mass of oxygen gas required to produce 65.75 grams of steam is approximately 162.2 grams

Explanation:

From the question, we have the following chemical reaction equation;

2C₃H₁₈(l) + 25O₂ (g) → 16CO₂(g) + 18H₂O (g)

The molar mass of oxygen, O₂ = 32 g/mol

The molar mass of steam, H₂O = 18.01528 g/mol

25 moles of oxygen are required to produce 18 moles of steam

Therefore, according to  Proust's law of definite proportions;

(32 × 25) g of oxygen are required to produce (18 × 18.01528) g of steam

65.75 g of steam will be produced by (32 × 25)/(18 × 18.01528) × 65.75 g ≈ 162.2 g of oxygen O₂.

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