Question

consider the unbalanced equation for the oxidation of butene. c4h8 + 6o2 → co2 + h2o for each molecule of c4h8 that reacts, how many molecules of carbon dioxide and water are produced? two carbon dioxide molecules and two water molecules four carbon dioxide molecules and four water molecules two carbon dioxide molecules and four water molecules four carbon dioxide molecules and two water molecules

Explanation:

Step1: Balance the carbon atoms

For 1 molecule of $C_4H_8$, to balance carbon in $CO_2$, we need 4 $CO_2$ as there are 4 carbon atoms in $C_4H_8$. The equation becomes $C_4H_8 + 6O_2
ightarrow4CO_2 + H_2O$.

Step2: Balance the hydrogen atoms

There are 8 hydrogen atoms in $C_4H_8$. In $H_2O$, to balance hydrogen, we need 4 $H_2O$ molecules since each $H_2O$ has 2 hydrogen atoms. The balanced equation is $C_4H_8+6O_2
ightarrow4CO_2 + 4H_2O$. So for each molecule of $C_4H_8$ that reacts, 4 carbon - dioxide and 4 water molecules are produced.

Answer:

B. four carbon dioxide molecules and four water molecules