Question

question 11 of 30
what is the balanced form of the chemical equation shown below?
c6h12o6(s) + o2(g) → h2o(l) + co2(g)
a. c6h12o6(s) + o2(g) → h2o(l) + co2(g)
b. ch2o(s) + o2(g) → h2o(l) + co2(g)
c. c6h12o6(s) + 6o2(g) → 6h2o(l) + 6co2(g)
d. c6h12o6(s) + o2(g) → 12h2o(l) + 6co2(g)

Explanation:

Step1: Balance carbon atoms

In $C_6H_{12}O_6$, there are 6 carbon atoms. On the right - hand side, $CO_2$ contains carbon. To balance the carbon atoms, we need 6 moles of $CO_2$. So the equation becomes $C_6H_{12}O_6(s)+O_2(g)
ightarrow H_2O(l)+6CO_2(g)$.

Step2: Balance hydrogen atoms

In $C_6H_{12}O_6$, there are 12 hydrogen atoms. On the right - hand side, $H_2O$ contains hydrogen. To balance the hydrogen atoms, we need 6 moles of $H_2O$ since each $H_2O$ has 2 hydrogen atoms. So the equation is now $C_6H_{12}O_6(s)+O_2(g)
ightarrow 6H_2O(l)+6CO_2(g)$.

Step3: Balance oxygen atoms

On the left - hand side, in $C_6H_{12}O_6$ there are 6 oxygen atoms. On the right - hand side, in 6 moles of $H_2O$ there are 6 oxygen atoms and in 6 moles of $CO_2$ there are 12 oxygen atoms, for a total of 18 oxygen atoms on the right - hand side. So we need 6 moles of $O_2$ on the left - hand side to balance the oxygen atoms. The balanced equation is $C_6H_{12}O_6(s)+6O_2(g)
ightarrow 6H_2O(l)+6CO_2(g)$.

Answer:

C. $C_6H_{12}O_6(s)+6O_2(g)
ightarrow 6H_2O(l)+6CO_2(g)$