Question

balance the reaction and do not leave any fractions.
c₆h₁₀ + o₂ → co₂ + h₂o
the guidelines in chapter 12 indicate that we should balance the o₂ last because it is an element, and any number placed in front of an element changes only that element. also, doing h and o last means that we should first balance the c. the are 6 on the left and 1 on the right so we put a 6 in front of the co₂. now the c is balanced.
c₆h₁₀ + o₂ → 6 co₂ + h₂o
we balance h next because o₂ is an element. there are 10 on the left and 2 on the right so we put a 1 in front of the h₂o. now the h is balanced.
c₆h₁₀ + o₂ → 2 co₂ + 3 h₂o
now, there are 17 o atoms on the right and only 2 on the left. on the left, we want 17 o atoms while each o₂ has 2. so, we put 17/2 in front of the o₂ which gives 8.5 molecules of o₂. the reaction is now balanced.
c₆h₁₀ + 17/2 o₂ → 4 co₂ + 5 h₂o
since there cannot be half of an o₂ molecule, we multiply the entire balanced reaction by 2 to give the final balanced reaction
6 c₆h₁₀ + 7 o₂ → 8 co₂ + 9 h₂o
a. 1 b. 2 c. 3 d. 4 e. 5 f. 6 g. 7 h. 8

Explanation:

Step1: Balance hydrogen atoms

On the left - hand side, there are 10 H atoms in \(C_{6}H_{10}\). In \(H_{2}O\), each molecule has 2 H atoms. To balance H, if we let the coefficient of \(H_{2}O\) be \(x\), then \(2x = 10\), so \(x = 5\).

Step2: Balance oxygen atoms after multiplying the entire equation by 2

The un - multiplied balanced equation (with a fraction for \(O_{2}\)) is \(C_{6}H_{10}+\frac{17}{2}O_{2}
ightarrow6CO_{2} + 5H_{2}O\). When we multiply the entire equation by 2:

• For \(C_{6}H_{10}\), the coefficient becomes \(2\) (because \(1\times2\)).
• For \(O_{2}\), \(\frac{17}{2}\times2=17\).
• For \(CO_{2}\), \(6\times2 = 12\).
• For \(H_{2}O\), \(5\times2=10\).

Answer:

1. E. 5
2. F. 6
3. E. 5
4. F. 6
5. E. 5
6. B. 2
7. G. 17
8. H. 12
9. H. 10