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What coefficient would the #O_2# have after balancing #C_2H_8 + O_2 -> CO_2 + H_2O#?

Answer 1

Natalie Anton

#C_2H_8+O_2rarrCO_2+H_2O#

First, I advise you to make a brief list of values for each side:

LHS

C = 2
H = 8 O = 2

RHS

C = 1 H = 2 O = 3

When balancing chemical equations, we should always try to balance the term with the most elements involved or one with an odd number of atoms. A single element is usually left until the end because it can be balanced without affecting other elements.

In this case, the oxygen on the RHS has only 3. If we multiply by 4, we are able to balance the hydrogen and get the oxygen even.

#C_2H_8+O_2rarrCO_2+4H_2O#

LHS

C = 2
H = 8 O = 2

RHS

C = 1 H = 8 O = 6

Let's multiply the #CO_2# 2 to balance to carbon:

#C_2H_8+O_2rarr2CO_2+4H_2O#

LHS

C = 2
H = 8 O = 2

RHS

C = 2 H = 8 O = 8

Finally, if we balance the oxygen by multiplying the #O_2# by 4, we are done:

#C_2H_8+4O_2rarr2CO_2+4H_2O#

Answer = 4

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Answer 2

Avery Adams

The coefficient for O2 after balancing the chemical equation C2H8 + 5O2 -> 2CO2 + 4H2O would be 5.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

What coefficient would the #O_2# have after balancing #C_2H_8 + O_2 -&gt; CO_2 + H_2O#?

What coefficient would the #O_2# have after balancing #C_2H_8 + O_2 -> CO_2 + H_2O#?