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 (c₆h₁₂o₆) reacts with 6 molecules of oxygen gas (o₂), the atoms in those molecules are rearranged to produce 6 molecules of carbon dioxide gas (co₂) and 6 molecules of water (h₂o). which chemical equation is balanced and represents the overall process of cellular respiration? 6cho + o → 6co + ho 6cho + 6o → 6co + 6ho cho + 6o → co + 6ho cho + 6o → 6co + 6ho

Explanation:

Step1: Count atoms in reactants

For glucose \(C_6H_{12}O_6\) (1 molecule) and \(O_2\) (6 molecules):

• Carbon: \(6\) atoms (from glucose).
• Hydrogen: \(12\) atoms (from glucose).
• Oxygen: \(6 + 6\times2=18\) atoms.

Step2: Count atoms in products

For \(CO_2\) (6 molecules) and \(H_2O\) (6 molecules):

• Carbon: \(6\times1 = 6\) atoms.
• Hydrogen: \(6\times2=12\) atoms.
• Oxygen: \(6\times2+6\times1 = 18\) atoms.

Step3: Analyze each equation

• In \(6CHO+O

ightarrow6CO + HO\): Atoms are not balanced (e.g., hydrogen and oxygen counts are wrong).

• In \(6CHO_2+6O

ightarrow6CO+6HO\): Formula of glucose is wrong (\(C_6H_{12}O_6\) is not represented correctly).

• In \(CHO+6O

ightarrow CO+6HO\): Carbon count is wrong (1 vs 6).

• In \(CHO_6+6O_2

ightarrow6CO_2+6HO_2\) (assuming \(CHO_6\) is \(C_6H_{12}O_6\) written in a non - standard way for the problem's options):

• Carbon: \(6\) (from \(C_6H_{12}O_6\)) and in products \(6\times1\) (from \(CO_2\)).
• Hydrogen: \(12\) (from \(C_6H_{12}O_6\)) and \(6\times2\) (from \(H_2O\)).
• Oxygen: \(6 + 6\times2=18\) (from \(C_6H_{12}O_6\) and \(O_2\)) and \(6\times2+6\times1 = 18\) (from \(CO_2\) and \(H_2O\)).

Answer:

\(CHO_6+6O_2
ightarrow6CO_2+6HO_2\) (assuming non - standard formula writing for the problem's options where \(CHO_6\) represents \(C_6H_{12}O_6\))