Answer:
There would be produced 0.00691 g of water and 0.034 g of CO₂
Explanation:
Combustion for benzene is:
2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O
2 moles of benzene react with 15 moles of oxygen to produce 6 moles of water and 12 moles of carbon dioxide.
Let's convert the mg to moles; previously we should convert mg to g
10 mg = 0.010 g
0.01 g / 78 g/m = 1.28×10⁻⁴ moles
Let's identify ratios. Ratio with CO₂ is 2:12 and 2:6 with water
2 moles of benzene react to produce 12 moles of CO₂
Then 1.28×10⁻⁴ mol of C₆H₆ would produce (1.28×10⁻⁴ . 12)/ 2 = 7.69×10⁻⁴ moles of CO₂.
2 moles of benzene react to produce 6 moles of water
Then 1.28×10⁻⁴ mol of C₆H₆ would produce (1.28×10⁻⁴ . 6) /2 = 3.84×10⁻⁴ moles of water
Let's convert the moles to mass (mol . molar mass)
7.69×10⁻⁴ moles of CO₂ . 44g/m = 0.034 g
3.84×10⁻⁴ moles of water . 18 g/m = 0.00691 g
A. The mass of carbon dioxide (CO₂) produced is 33.84 mg
B. The mass of water (H₂O) produced is 6.92 mg
2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O
Molar mass of C₆H₆ = (12×6) + (1×6) = 78 g/mol
Mass of C₆H₆ from the balanced equation = 2 × 78 = 156 g
Molar mass of CO₂ = 12 + (16×2) = 44 g/mol
Mass of CO₂ from the balanced equation = 12 × 44 = 528 g
Molar mass of H₂O = (2×1) + 16 = 18 g/mol
Mass of H₂O from the balanced equation = 6 × 18 = 108 g
SUMMARY
From the balanced equation above,
156 g of C₆H₆ reacted to produce 528 g of CO₂ and 108 g of H₂O
From the balanced equation above,
156 g of C₆H₆ reacted to produce 528 g of CO₂.
Therefore,
10 mg of C₆H₆ will react to produce = (10 × 528) / 156 = 33.84 mg of CO₂.
Thus, 33.84 mg of CO₂ were produced from the reaction.
From the balanced equation above,
156 g of C₆H₆ reacted to produce 108 g of H₂O.
10 mg of C₆H₆ will react to produce = (10 × 108) / 156 = 6.92 mg of H₂O
Thus, 6.92 mg of H₂O were produced from the reaction.
Learn more about stoichiometry:
https://brainly.com/question/14735801
