Answer:
4.58g of CO₂ could be produced
Explanation:
Based on the reaction:
C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(g)
<em>1 mole of C₂H₄ reacts with 3 moles of oxygen to produce 2 moles of CO₂</em>
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To solve this question we must find the moles of each reactant in order to find limiting reactant. With limiting reactant we can find the moles -And the mass- of CO₂ produced:
<em>Moles C₂H₄ -Molar mass: 28.05g/mol-</em>
2.0g * (1mol / 28.05g) = 0.0713moles
<em>Moles O₂ -Molar mass: 32g/mol-</em>
5.0g * (1mol / 32g) = 0.156moles
For a complete reaction of 0.0713 moles of C2H4 are required:
0.0713 moles C₂H₄ * (3 moles O₂ / 1 mol C₂H₄) = 0.214 moles of O₂
As there are just 0.156 moles, O₂ is limiting reactant.
The moles of CO₂ produced are:
0.156 moles O₂ * (2mol CO₂ / 3mol O₂) = 0.104 moles CO₂
The mass is -Molar mass CO₂: 44.01g/mol-
0.104 moles CO₂ * (44.01g / mol) =
<h3>4.58g of CO₂ could be produced</h3>
