Question

question 17
1 pts
when the following reaction is balanced the coefficient in front of carbon dioxide is:
__ c5h12 + o2 → co2 + __ h2o
6
10
12
5
question 18
1 pts
when the following reaction is balanced the coefficient in front of oxygen is:
__ c8h18 + o2 → co2 + __ h2o
18
9
25
12

Explanation:

Step1: Balance the carbon atoms

For the reaction \(C_{5}H_{12}+O_{2}\to CO_{2}+H_{2}O\), there are 5 carbon atoms in \(C_{5}H_{12}\). So, we put a coefficient of 5 in front of \(CO_{2}\) to balance the carbon atoms. The equation becomes \(C_{5}H_{12}+O_{2}\to5CO_{2}+H_{2}O\).

Step2: Balance the hydrogen atoms

There are 12 hydrogen atoms in \(C_{5}H_{12}\). So, we put a coefficient of 6 in front of \(H_{2}O\) to balance the hydrogen atoms. The equation becomes \(C_{5}H_{12}+O_{2}\to5CO_{2}+6H_{2}O\).

Step3: Balance the oxygen atoms

On the right - hand side, there are \(5\times2 + 6\times1=10 + 6 = 16\) oxygen atoms. So, we put a coefficient of 8 in front of \(O_{2}\) to balance the oxygen atoms. The balanced equation is \(C_{5}H_{12}+8O_{2}\to5CO_{2}+6H_{2}O\).

For the reaction \(C_{8}H_{18}+O_{2}\to CO_{2}+H_{2}O\):

Step1: Balance the carbon atoms

There are 8 carbon atoms in \(C_{8}H_{18}\). So, we put a coefficient of 8 in front of \(CO_{2}\). The equation becomes \(C_{8}H_{18}+O_{2}\to8CO_{2}+H_{2}O\).

Step2: Balance the hydrogen atoms

There are 18 hydrogen atoms in \(C_{8}H_{18}\). So, we put a coefficient of 9 in front of \(H_{2}O\). The equation becomes \(C_{8}H_{18}+O_{2}\to8CO_{2}+9H_{2}O\).

Step3: Balance the oxygen atoms

On the right - hand side, there are \(8\times2+9\times1 = 16 + 9=25\) oxygen atoms. So, we put a coefficient of \(\frac{25}{2}\) in front of \(O_{2}\). To get rid of the fraction, we multiply the entire equation by 2: \(2C_{8}H_{18}+25O_{2}\to16CO_{2}+18H_{2}O\).

Answer:

For the first reaction (Question 17), the coefficient in front of \(CO_{2}\) is 5.
For the second reaction (Question 18), the coefficient in front of \(O_{2}\) is 25.