consider the chemical equation for the combustion of sugar.\n_$c_{6}h_{12}o_{6}(s)$ + _$o_{2}(g)$…

consider the chemical equation for the combustion of sugar.\n_$c_{6}h_{12}o_{6}(s)$ + _$o_{2}(g)$ $\rightarrow$ _$co_{2}(g)$ + _$h_{2}o(l)$\nwhich sequence of coefficients should be placed in the blanks to balance this equation?\no 1, 6, 6, 6\no 6, 1, 6, 1\no 3, 3, 3, 6\no 1, 3, 3, 6

consider the chemical equation for the combustion of sugar.\n_$c_{6}h_{12}o_{6}(s)$ + _$o_{2}(g)$ $\rightarrow$ _$co_{2}(g)$ + _$h_{2}o(l)$\nwhich sequence of coefficients should be placed in the blanks to balance this equation?\no 1, 6, 6, 6\no 6, 1, 6, 1\no 3, 3, 3, 6\no 1, 3, 3, 6

Answer

Explanation:

Step1: Balance carbon atoms

On the left - hand side, in $C_6H_{12}O_6$, there are 6 carbon atoms. In $CO_2$ on the right - hand side, to balance the carbon atoms, if we put a coefficient of 6 in front of $CO_2$, we have 6 carbon atoms on both sides. The equation becomes $C_6H_{12}O_6(s)+O_2(g)\to6CO_2(g)+H_2O(l)$.

Step2: Balance hydrogen atoms

In $C_6H_{12}O_6$, there are 12 hydrogen atoms. In $H_2O$ on the right - hand side, to balance the hydrogen atoms, if we put a coefficient of 6 in front of $H_2O$, we have 12 hydrogen atoms on both sides. The equation is now $C_6H_{12}O_6(s)+O_2(g)\to6CO_2(g)+6H_2O(l)$.

Step3: Balance oxygen atoms

On the right - hand side, in $6CO_2$ there are $6\times2 = 12$ oxygen atoms and in $6H_2O$ there are 6 oxygen atoms, so a total of $12 + 6=18$ oxygen atoms. In $C_6H_{12}O_6$ there are 6 oxygen atoms. So, the number of oxygen atoms from $O_2$ needed is $18 - 6=12$. Since each $O_2$ molecule has 2 oxygen atoms, we need 6 $O_2$ molecules. The balanced equation is $C_6H_{12}O_6(s)+6O_2(g)\to6CO_2(g)+6H_2O(l)$.

Answer:

1, 6, 6, 6