Question

which chemical reaction involves the fewest oxygen atoms?
○ ( 2\text{agno}_3 + \text{k}_2\text{so}_4
ightarrow \text{ag}_2\text{so}_4 + 2\text{kno}_3 )
○ ( 6\text{co}_2 + 6\text{h}_2\text{o}
ightarrow \text{c}_6\text{h}_{12}\text{o}_6 + 6\text{o}_2 )
○ ( \text{c}_2\text{h}_4 + 3\text{o}_2
ightarrow 2\text{co}_2 + 2\text{h}_2\text{o} )
○ ( 4\text{fe} + 6\text{h}_2\text{o} + 3\text{o}_2
ightarrow 4\text{fe(oh)}_3 )

Explanation:

Step1: Calculate oxygen atoms in first reaction

For \(2\text{AgNO}_3 + \text{K}_2\text{SO}_4
ightarrow \text{Ag}_2\text{SO}_4 + 2\text{KNO}_3\):
Oxygen in reactants: \(2\times3 + 4 = 10\) (since \(2\text{AgNO}_3\) has \(2\times3\) O, \(\text{K}_2\text{SO}_4\) has 4 O).
Oxygen in products: \(4 + 2\times3 = 10\) (same as reactants, conservation).

Step2: Calculate oxygen atoms in second reaction

For \(6\text{CO}_2 + 6\text{H}_2\text{O}
ightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\):
Oxygen in reactants: \(6\times2 + 6\times1 = 18\).
Oxygen in products: \(6 + 6\times2 = 18\) (conservation).

Step3: Calculate oxygen atoms in third reaction

For \(\text{C}_2\text{H}_4 + 3\text{O}_2
ightarrow 2\text{CO}_2 + 2\text{H}_2\text{O}\):
Oxygen in reactants: \(3\times2 = 6\).
Oxygen in products: \(2\times2 + 2\times1 = 6\) (conservation).

Step4: Calculate oxygen atoms in fourth reaction

For \(4\text{Fe} + 6\text{H}_2\text{O} + 3\text{O}_2
ightarrow 4\text{Fe(OH)}_3\):
Oxygen in reactants: \(6\times1 + 3\times2 = 12\).
Oxygen in products: \(4\times3 = 12\) (conservation).

Answer:

The reaction \(\boldsymbol{\text{C}_2\text{H}_4 + 3\text{O}_2
ightarrow 2\text{CO}_2 + 2\text{H}_2\text{O}}\) involves the fewest oxygen atoms (6 total).