QUESTION IMAGE
Question
consider the chemical equation for the combustion of sugar. _c6h12o6(s) + _o2(g) → _co2(g) + _h2o(l) which sequence of coefficients should be placed in the blanks to balance this equation? 1, 6, 6, 6 6, 1, 6, 1 3, 3, 3, 6 1, 3, 3, 6
Step1: Balance carbon atoms
On the left - hand side, there are 6 carbon atoms in $C_{6}H_{12}O_{6}$. So, we put a 6 in front of $CO_{2}$ on the right - hand side to balance carbon: $C_{6}H_{12}O_{6}(s)+O_{2}(g)\to6CO_{2}(g)+H_{2}O(l)$.
Step2: Balance hydrogen atoms
There are 12 hydrogen atoms in $C_{6}H_{12}O_{6}$. So, we put a 6 in front of $H_{2}O$ on the right - hand side to balance hydrogen: $C_{6}H_{12}O_{6}(s)+O_{2}(g)\to6CO_{2}(g)+6H_{2}O(l)$.
Step3: Balance oxygen atoms
On the right - hand side, the total number of oxygen atoms in $6CO_{2}$ and $6H_{2}O$ is $6\times2 + 6\times1=18$. In $C_{6}H_{12}O_{6}$, there are 6 oxygen atoms. So, we need $18 - 6=12$ more oxygen atoms from $O_{2}$. Since each $O_{2}$ molecule has 2 oxygen atoms, we put a 6 in front of $O_{2}$ on the left - hand side: $C_{6}H_{12}O_{6}(s)+6O_{2}(g)\to6CO_{2}(g)+6H_{2}O(l)$.
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