For the following reaction, 8.54 grams of carbon monoxide are allowed to react with 7.91 grams of oxygen gas .
carbon monoxide(g) + oxygen(g) carbon dioxide(g)
What is the maximum mass of carbon dioxide that can be formed?
grams
What is the FORMULA for the limiting reagent?
What mass of the excess reagent remains after the reaction is complete?
grams
Chemistry?
2017-11-08 12:16 pm
回答 (1)
2017-11-08 4:34 pm
Molar mass of CO = (12.0 + 16.0) g/mol = 28.0 g/mol
Molar mass of O₂ = 16.0 × 2 = 32.0 g/mol
Molar mass of CO₂ = (12.0 + 16.0×2) g/mol = 44.0 g/mol
Initial number of moles of CO = (8.54 g) / (28.0 g/mol) = 0.305 mol
Initial number of moles of O₂ = (7.91 g) / (32.0 g/mol) = 0.247 mol
Balanced equation for the reaction :
2CO(g) + O₂(g) → 2CO₂(g)
Mole ratio CO : O₂ : CO₂ = 2 : 1 : 2
When 0.305 mole of CO completely reacted, O₂ needed = (0.305 mol) × (1/2) = 0.1525 mol < Initial number of moles of O₂
Hence, O₂ is in excess, and the limiting reagent is CO.
Number of moles of CO reacted = 0.305 mol
Maximum number of moles of CO₂ formed = (0.305 mol) × (2/2) = 0.305 mol
Maximum mass of CO₂ formed = (0.305 mol) × (44.0 g/mol) = 13.4 g
Number of moles of O₂ reacted = (0.305 mol) × (1/2) = 0.305/2 mol
Mass of O₂ reacted = (0.305/2 mol) × (32 g/mol) = 4.88 g
Mass of excess O₂ after the reaction is complete = (7.91 - 4.88) g = 3.03 g
Molar mass of O₂ = 16.0 × 2 = 32.0 g/mol
Molar mass of CO₂ = (12.0 + 16.0×2) g/mol = 44.0 g/mol
Initial number of moles of CO = (8.54 g) / (28.0 g/mol) = 0.305 mol
Initial number of moles of O₂ = (7.91 g) / (32.0 g/mol) = 0.247 mol
Balanced equation for the reaction :
2CO(g) + O₂(g) → 2CO₂(g)
Mole ratio CO : O₂ : CO₂ = 2 : 1 : 2
When 0.305 mole of CO completely reacted, O₂ needed = (0.305 mol) × (1/2) = 0.1525 mol < Initial number of moles of O₂
Hence, O₂ is in excess, and the limiting reagent is CO.
Number of moles of CO reacted = 0.305 mol
Maximum number of moles of CO₂ formed = (0.305 mol) × (2/2) = 0.305 mol
Maximum mass of CO₂ formed = (0.305 mol) × (44.0 g/mol) = 13.4 g
Number of moles of O₂ reacted = (0.305 mol) × (1/2) = 0.305/2 mol
Mass of O₂ reacted = (0.305/2 mol) × (32 g/mol) = 4.88 g
Mass of excess O₂ after the reaction is complete = (7.91 - 4.88) g = 3.03 g
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