CHEMISTRY | MOLE CONCEPT| Question 15
If 4.95g 0f ethylene (C2H4) are combusted with 3.25g of Oxygen
a) What is the limiting reagent?
b) How many gram of CO2 are formed?
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DATA GIVEN:
Mass of ethylene (C2H4) = 4.95g
Mass of oxygen (O2) = 3.25g
Molar mass of C2H4 = 28g/mole
Molar mass of O2 = 32g/mole
Mole for chemical equation
C2H4 + 302 → 2CO2 + 2H2> O
1 mole: 3 moles: 2 moles: 2 moles
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Mole ratio
To calculate mole for ethylene (C2H4) = mass / Molar mass
= 4.95g / 28 g/mole
= 0.176785 mole
Mole ratio for C2H4 = 0.176785 mole / 1 mole
= 0.176785
For oxygen (O2) = Mass / Molar mass
= 3.25g / 32g/mole
= 0.1015625 mole
Mole ratio = 0.1015625 moles / 3 moles
= 0.03385
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b/ To find the mass of CO2 formed
From mole equivalent between O2 and CO2
= (0.1015625 mole × 2 moles) / 3 moles
= 0.06777083333 moles
Mole = mass / Molar mass
Mass = mole × molar mass of CO2
= 0.067708 × 44
= 2.9791g
2.9791g of carbon dioxide was formed
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c/ The limiting reagent
The limiting reagent is oxygen (O2)
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