Question

consider the unbalanced equation for the oxidation of butene.
$c_4h_8 + 6o_2 \to co_2 + h_2o$
for each molecule of $c_4h_8$ that reacts, how many molecules of carbon dioxide and water are produced?
four carbon dioxide
molecules and four
water molecules
two carbon dioxide
molecules and two water
molecules
two carbon dioxide
molecules and four
water molecules
four carbon dioxide
molecules and two water
molecules

Explanation:

Step1: Balance the carbon atoms

En la ecuación no balanceada \(C_{4}H_{8}+6O_{2}\to CO_{2}+H_{2}O\), hay 4 átomos de carbono en \(C_{4}H_{8}\). Entonces, debemos poner un coeficiente 4 frente a \(CO_{2}\) para balancear los átomos de carbono: \(C_{4}H_{8}+6O_{2}\to4CO_{2}+H_{2}O\).

Step2: Balance the hidrógeno atoms

Hay 8 átomos de hidrógeno en \(C_{4}H_{8}\). Entonces, debemos poner un coeficiente 4 frente a \(H_{2}O\) para balancear los átomos de hidrógeno: \(C_{4}H_{8}+6O_{2}\to4CO_{2}+4H_{2}O\).

Step3: Verificar el balance de oxígeno

En el lado izquierdo, hay \(6\times2 = 12\) átomos de oxígeno en \(O_{2}\). En el lado derecho, en \(4CO_{2}\) hay \(4\times2=8\) átomos de oxígeno y en \(4H_{2}O\) hay \(4\times1 = 4\) átomos de oxígeno. En total, \(8 + 4=12\) átomos de oxígeno. La ecuación balanceada es \(C_{4}H_{8}+6O_{2}=4CO_{2}+4H_{2}O\).

Answer:

four carbon dioxide molecules and four water molecules