The mass of water that will be produced if 10.54 g of H₂ react with 95.10 g of O₂ is 94.86 g
Balanced equation
2H₂ + O₂ —> 2H₂O
Molar mass of H₂ = 2 × 1 = 2 g/mol
Mass of H₂ from the balanced equation = 2 × 2 = 4 g
Molar mass of O₂ = 2 × 16 = 32 g/mol
Mass of O₂ from the balanced equation = 1 × 32 = 32 g
Molar mass of H₂O = (2×1) + 16 = 18 g/mol
Mass of H₂O from the balanced equation = 2 ×18 = 36 g
SUMMARY
From the balanced equation above,
4 g of H₂ reacted with 32 g of O₂ to produce 36 g of H₂O
How to determine the limiting reactant
From the balanced equation above,
4 g of H₂ reacted with 32 g of O₂
Therefore,
10.54 g of H₂ will react with = (10.54 × 32) / 4 = 84.32 g of O₂
From the calculation made above, we can see that only 84.32 g out of 95.10 g of O₂ given, is needed to react completely with 10.54 g of H₂.
Therefore, H₂ is the limiting reactant.
How to determine the mass of water produced
From the balanced equation above,
4 g of H₂ reacted to produce 36 g of H₂O
Therefore,
10.54 g of H₂ will react to produce = (10.54 × 36) / 4 = 94.86 g of H₂O
Thus, 94.86 g of H₂O were obtained from the reaction.
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