Question

kcw writing and balancing chemical equations
me balancing chemical equations: read sections 7.3, 7.6.
you can click on the review link to access the section in your
text
complete and balance each of the following combustion
reaction equations.
part b
c4h10(g)+o2(g)→
express your answer as a complete chemical equation including phases.
a chemical reaction does not occur for this question

Explanation:

Step1: Balance carbon atoms

Carbon atoms are present in \(C_{4}H_{10}\) and \(CO_{2}\). In \(C_{4}H_{10}\), there are 4 carbon atoms. So, we put a coefficient of 4 in front of \(CO_{2}\).
\(C_{4}H_{10}(g)+O_{2}(g)
ightarrow4CO_{2}(g)+H_{2}O(l)\)

Step2: Balance hydrogen atoms

Hydrogen atoms are in \(C_{4}H_{10}\) and \(H_{2}O\). In \(C_{4}H_{10}\), there are 10 hydrogen atoms. So, we put a coefficient of 5 in front of \(H_{2}O\).
\(C_{4}H_{10}(g)+O_{2}(g)
ightarrow4CO_{2}(g)+5H_{2}O(l)\)

Step3: Balance oxygen atoms

Oxygen atoms are in \(O_{2}\), \(CO_{2}\), and \(H_{2}O\). On the right - hand side, the number of oxygen atoms is \(4\times2 + 5\times1=8 + 5=13\). So, we put a coefficient of \(\frac{13}{2}\) in front of \(O_{2}\). But we usually prefer whole - number coefficients. Multiply the entire equation by 2 to get rid of the fraction.
\(2C_{4}H_{10}(g)+13O_{2}(g)
ightarrow8CO_{2}(g)+10H_{2}O(l)\)

Answer:

\(2C_{4}H_{10}(g)+13O_{2}(g)
ightarrow8CO_{2}(g)+10H_{2}O(l)\)