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. If sulphur (IV) oxide and methane are
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The mass of water (H₂O) that would be produced is 4.5 g
<h3>Stoichiometry </h3>From the question, we are to determine the mass of water that would be produced.
From the given balanced chemical equation
2C₂H₆ +7O₂ → 4CO₂ +6H₂O
This means
2 moles of C₂H₆ reacts with 7 moles of O₂ to produce 4 moles of CO₂ and 6 moles of H₂O
Now, we will determine the number of moles of each reactant present
- For Ethane (C₂H₆)
Mass = 2.5 g
Molar mass = 30.07 g
Using the formula,
Number of moles of C₂H₆ present =
Number of moles of C₂H₆ present = 0.08314 mole
- For Oxygen (O₂)
Mass = 170g
Molar mass = 31.999 g/mol
Number of moles of O₂ present =
Number of moles of O₂ present = 5.3127 moles
Since
2 moles of C₂H₆ reacts with 7 moles of O₂
Then,
0.08314 mole of C₂H₆ will react with
= 0.58198 mole
Therefore,
0.08314 mole of C₂H₆ reacts with 0.58198 mole of O₂ to produce 3 × 0.08314 moles of H₂O
3 × 0.08314 = 0.24942 mole
Thus, the number of moles of water (H₂O) produced is 0.24942 mole
Now, for the mass of water that would be produced,
Using the formula,
Mass = Number of moles × Molar mass
Molar mass of water = 18.015 g/mol
Then,
Mass of water that would be produced = 0.24942 × 18.015
Mass of water that would be produced = 4.4933 g
Mass of water that would be produced ≅ 4.5 g
Hence, the mass of water (H₂O) that would be produced is 4.5 g
Learn more on Stoichiometry here: brainly.com/question/14271082