Question

How many moles of O2 are required for the complete reaction of 88.9 g of C2H4 to form CO2 and H2O?

Answer 1

First, you need to right the balance equation:

C2H4 + 3O2 --> 2H2O + 2CO2

Then you can use stoichoimestry to solve.

88.9 g C2H4 (1 mole/ 28.04 g) (3 mole O2/1 mil C2H4)= 9.51 mole of O2

Answer 2

9.525 moles of O₂

Further explanation

Given:

Combustion of 88.9 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.

$\boxed{ \ n = \frac{mass}{Mr} \ } \rightarrow \boxed{ \ n = \frac{88.9}{28} = 3.175 \ moles \ }$

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.

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

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

$\boxed{ \ n(O_2) = \frac{3}{1} \times 3.175 \ moles \ }$

Thus, the number of moles of O are required for the complete reaction of the combustion of C₂H₄ is 9.525 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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