27.0 grams of hydrogen and 27.0 grams of oxygens are reacted to form water Oxygen is limiting reactant
What mass of the excess reactant remains after the reaction is over
2) how many grams of water is produced?
chem please Please use bca table?
2017-01-13 9:51 pm
回答 (1)
2017-01-13 10:12 pm
1)
Molar mass of H₂ = 1.0 × 2 g/mol = 2.0 g/mol
Molar mass of O₂ = 16.0 × 2 g/mol = 32.0 g/mol
2H₂ + O₂ → 2H₂O
OR: Mole ratio H₂ : O₂ = 2 : 1
No. of moles of O₂ reacted = (27.0 g) / (32.0 g/mol) = 0.844 mol
No. of moles of H₂ reacted = (0.844 mol) × 2 = 1.69 mol
Mass of H₂ reacted = (1.69 mol) × (2.0 g/mol) = 3.4 g
Mass of the excess reactant (H₂) remains = (27.0 - 3.4) g = 23.6 g
2)
Molar mass of H₂O = (1.0×2 + 16.0) g/mol = 18.0 g/mol
According to the equation in 1), mole ratio O₂ : H₂O = 1 : 2
No. of moles of O₂ reacted = 0.844 mol
No. of moles of H₂O produced = (0.844 mol) × 2 = 1.69 mol
Mass of H₂O produced = (1.69 mol) × (18.0 g/mol) = 30.4 g
Molar mass of H₂ = 1.0 × 2 g/mol = 2.0 g/mol
Molar mass of O₂ = 16.0 × 2 g/mol = 32.0 g/mol
2H₂ + O₂ → 2H₂O
OR: Mole ratio H₂ : O₂ = 2 : 1
No. of moles of O₂ reacted = (27.0 g) / (32.0 g/mol) = 0.844 mol
No. of moles of H₂ reacted = (0.844 mol) × 2 = 1.69 mol
Mass of H₂ reacted = (1.69 mol) × (2.0 g/mol) = 3.4 g
Mass of the excess reactant (H₂) remains = (27.0 - 3.4) g = 23.6 g
2)
Molar mass of H₂O = (1.0×2 + 16.0) g/mol = 18.0 g/mol
According to the equation in 1), mole ratio O₂ : H₂O = 1 : 2
No. of moles of O₂ reacted = 0.844 mol
No. of moles of H₂O produced = (0.844 mol) × 2 = 1.69 mol
Mass of H₂O produced = (1.69 mol) × (18.0 g/mol) = 30.4 g
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