Question

What coefficient is placed in front of O2 to complete the balancing of the following equation?C5H8 (1)+7 02 (g) - 5 CO2 (g) + 4H2O(g)

Answer 1

Answer:

The coefficient in front of O2 is 7.

Explanation:

To balance the chemical equation, we have to make sure that there is the same amount of each element in the reactants side (to the left of the arrow) and in the products side (to the right of the arrow).

- The original equation is:

$C_5H_8+O_2\operatorname{\rightarrow}CO_2+H_2O$

And in the reactants side there are: 5 C, 8 H and 2 O.

While in the products side there are: 1 C, 2 H and 3 O. None of the elements are balanced.

- So, we first balance the number of carbons in the products side, adding 5, and the equation will be:

$C_5H_8+O_2\operatorname{\rightarrow}5CO_2+H_2O$

Here, in the reactants side there are: 5 C, 8 H and 2 O.

While in the products side there are: 5 C, 2 H and 11 O.

Only Carbon is balanced.

- Now, it is necessary to balance the hydrogens in the products side, by adding 4 in front of the water molecule:

$C_5H_8+O_2\operatorname{\rightarrow}5CO_2+4H_2O$

Here, in the reactants side there are: 5 C, 8 H and 2 O.

While in the products side there are: 5 C, 8 H and 14 O.

Carbon and hydrogen are balanced.

- Finally, to balance the oxygens in the reactants side, it is necessary to add 7, so 7 x 2 = 14 (7 is the coefficient and 2 is the subscript in O2):

$C_5H_8+7O_2\operatorname{\rightarrow}5CO_2+4H_2O$

Now, in both sides of the reaction there are: 5 C, 8 H and 14 O. The reaction is now balanced, and the coefficient in front of O2 is 7.