Question

How many moles of o2 are required for the complete reaction of 54.7 g of c2h4 to form co2 and h2o?

Answer 1

The balanced chemical equation that illustrates this reaction is:
C2H4 + 3O2 --> 2CO2 + 2H2O 

From the periodic table:
mass of carbon = 12 grams
mass of hydrogen = 1 gram
Therefore:
molar mass of C2H4 = 12(2) + 4(1) = 24 + 4 = 28 grams

number of moles = mass / molar mass
number of moles of C2H4 = 54.7 / 28 = 1.95 moles

From the balanced equation above:
3 moles of oxygen are required to react with one mole of C2H4, therefore, to know the number of moles required to react with 1.95 moles of C2H4, all you have to do is cross multiplication as follows:
number of oxygen moles = (1.95*3) / 1 = 5.85 moles

Answer 2

5.862 moles of O₂

Further explanation

Given:

Combustion of 54.7 g of C₂H₄ to form CO₂ and H₂O.

Question:

How many moles of O₂ are required for the complete reaction of combustion of C₂H₄?

The Process:

• Relative atomic mass: C = 12 and H = 1.
• Relative molecular mass (Mr) of C₂H₄ = 2(12) + 4(1) = 28.

Let us convert mass to mole for C₂H₄.

$\boxed{ \ n = \frac{mass}{Mr} \ } \rightarrow \boxed{ \ n = \frac{54.7}{28} = 1.954 \ moles \ }$

The combustion reaction of  C₂H₄ (ethylene, also named ethene) can be expressed as follows:

$\boxed{ \ C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O \ }$ (the reaction is balanced)

According to chemical equation above, proportion between C₂H₄ and O₂ is 1 to 3. Therefore, we can count the number of moles of O₂.

$\boxed{ \ \frac{n(O_2)}{n(C_2H_4)} = \frac{3}{1} \ }$

$\boxed{ \ n(O_2) = \frac{3}{1} \times n(C_2H_4) \ }$

$\boxed{ \ n(O_2) = \frac{3}{1} \times 1.954 \ moles \ }$

Thus, the number of moles of O are required for the complete reaction of the combustion of C₂H₄ is 5.862 moles.

_ _ _ _ _ _ _ _ _

Notes:

If we want to calculate the mass of O₂, then we use the number of moles of O₂ that have been obtained.

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