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Kobotan [32]
3 years ago
10

Balance this equation. If a coefficient of "1" is required, choose "blank" for that box. C2H6 + O2 → CO2 + H2O

Chemistry
2 answers:
Black_prince [1.1K]3 years ago
6 0

Explanation:

In a balanced equation, number of atoms on both reactant and product side are equal.

For example, C_{2}H_{6} + O_{2} \rightarrow CO_{2} + H_{2}O

Number of atoms present on reactant side are as follows.

C = 2

H = 6

O = 2

Number of atoms present on product side are as follows.

C = 1

H = 2

O = 3

Therefore, to balance this equation multiply C_{2}H_{6} by 2 and multiply O_{2} by 7 on reactant side. Whereas multiply CO_{2} by 4 and multiply H_{2}O by 6 on product side.

Therefore, the balanced chemical equation is as follows.

          2C_{2}H_{6} + 7O_{2} \rightarrow 4CO_{2} + 6H_{2}O

fenix001 [56]3 years ago
5 0
Step 1: Write the unbalanced equation,

                                 C₂H₆  +  O₂    →    CO₂  +  H₂<span>O

There are 2 C at left hand side and 1 carbon at right hand side. So, multiply CO</span>₂ by 2 to balance C atoms at both side. So,

                                 C₂H₆  +  O₂    →   2 CO₂  +  H₂O

Now, count number of H atoms at both sides. There are 6 H atoms at left hand side and 2 at right hand side. Multiply H₂O by 3 to balance H atoms.


                                 C₂H₆  +  O₂    →   2 CO₂  +  3 H₂O

At last, balance O atoms. There are 2 O atoms at left hand side and 3 O atoms at right hand side. Multiply O₂ with 1.5 (i.e. 3/2) to balance O atoms. i.e.

                                 C₂H₆  +  3/2 O₂    →   2 CO₂  +  3 H₂O

Hence, the equation is balanced. If you want to make equation fraction free then multiply all equation with 2. i.e.

                           ( C₂H₆  +  3/2 O₂    →   2 CO₂  +  3 H₂O ) × 2

                                2 C₂H₆  +  3 O₂    →   4 CO₂  +  6 H₂O
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In standardizing a naoh solution a student found that 25.55cm of base neutralize exactly 21.35cm of 0.12M HCl find the molarity
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Answer:

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Explanation:

First let's generate a balanced equation for the reaction

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From the equation,

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