Question

the reactants and products of cellular respiration are made of the same amounts of hydrogen, carbon, and oxygen atoms. when 1 molecule of glucose ($c_6h_{12}o_6$) reacts with 6 molecules of oxygen gas ($o_2$), the atoms in those molecules are rearranged to produce 6 molecules of carbon dioxide gas ($co_2$) and 6 molecules of water ($h_2o$).
which chemical equation is balanced and represents the overall process of cellular respiration?
$c_6h_{12}o_6 + 6o_2 \longrightarrow co_2 + 6h_2o$ $c_6h_{12}o_6 + 6o_2 \longrightarrow 6co_2 + 6h_2o$
$6c_6h_{12}o_6 + 6o_2 \longrightarrow 6co_2 + 6h_2o$ $6c_6h_{12}o_6 + o_2 \longrightarrow 6co_2 + h_2o$

Explanation:

Step1: Count C atoms on left

Left side: $C_6H_{12}O_6$ has 6 C atoms.

Step2: Match C atoms on right

Right side needs 6 C atoms, so $CO_2$ needs coefficient 6: $6CO_2$.

Step3: Count H atoms on left

Left side: $C_6H_{12}O_6$ has 12 H atoms.

Step4: Match H atoms on right

Right side: $H_2O$ has 2 H each. $\frac{12}{2}=6$, so $H_2O$ needs coefficient 6: $6H_2O$.

Step5: Count O atoms on right

Right side: $6CO_2$ has $6\times2=12$ O; $6H_2O$ has $6\times1=6$ O. Total O: $12+6=18$.

Step6: Match O atoms on left

Left side: $C_6H_{12}O_6$ has 6 O. Remaining O needed: $18-6=12$. $O_2$ has 2 O each. $\frac{12}{2}=6$, so $O_2$ needs coefficient 6: $6O_2$.

Step7: Verify balanced equation

Left: $6C, 12H, 6+(6\times2)=18O$; Right: $6C, 6\times2=12H, (6\times2)+6=18O$. All atoms balanced.

Answer:

$C_6H_{12}O_6 + 6O_2 \longrightarrow 6CO_2 + 6H_2O$