Question

nitroglycerin is a dangerous powerful explosive that violently decomposes when it the nobel prizes with a fortune he made by inventing dynamite, a mixture of nitro

1. write a balanced chemical equation, including physical state symbols, for the

decomposition of liquid nitroglycerin $(c_{3}h_{5}(no_{3})_{3})$ into gaseous dinitrogen,
gaseous dioxygen, gaseous water and gaseous carbon dioxide.

Explanation:

Step1: Write the un - balanced equation

The decomposition of liquid nitroglycerin \(C_{3}H_{5}(NO_{3})_{3}(l)\) gives gaseous dinitrogen \(N_{2}(g)\), gaseous dioxygen \(O_{2}(g)\), gaseous water \(H_{2}O(g)\) and gaseous carbon dioxide \(CO_{2}(g)\).
The un - balanced equation is:
\(C_{3}H_{5}(NO_{3})_{3}(l)
ightarrow N_{2}(g)+O_{2}(g)+H_{2}O(g)+CO_{2}(g)\)

Step2: Balance the atoms

• Carbon atoms:

On the left - hand side, the number of carbon atoms in \(C_{3}H_{5}(NO_{3})_{3}\) is \(3\). On the right - hand side, carbon is in \(CO_{2}\). So we put a coefficient of \(3\) in front of \(CO_{2}\).
The equation becomes \(C_{3}H_{5}(NO_{3})_{3}(l)
ightarrow N_{2}(g)+O_{2}(g)+H_{2}O(g)+3CO_{2}(g)\)

• Hydrogen atoms:

On the left - hand side, the number of hydrogen atoms in \(C_{3}H_{5}(NO_{3})_{3}\) is \(5\). On the right - hand side, hydrogen is in \(H_{2}O\). So we put a coefficient of \(\frac{5}{2}\) in front of \(H_{2}O\).
The equation becomes \(C_{3}H_{5}(NO_{3})_{3}(l)
ightarrow N_{2}(g)+O_{2}(g)+\frac{5}{2}H_{2}O(g)+3CO_{2}(g)\)

• Nitrogen atoms:

On the left - hand side, the number of nitrogen atoms in \(C_{3}H_{5}(NO_{3})_{3}\) is \(3\). On the right - hand side, nitrogen is in \(N_{2}\). So we put a coefficient of \(\frac{3}{2}\) in front of \(N_{2}\).
The equation becomes \(C_{3}H_{5}(NO_{3})_{3}(l)
ightarrow\frac{3}{2}N_{2}(g)+O_{2}(g)+\frac{5}{2}H_{2}O(g)+3CO_{2}(g)\)

• Oxygen atoms:

On the left - hand side, the number of oxygen atoms in \(C_{3}H_{5}(NO_{3})_{3}\) is \(9\).
On the right - hand side, the number of oxygen atoms:
In \(\frac{3}{2}N_{2}\), there are \(0\) oxygen atoms. In \(\frac{5}{2}H_{2}O\), the number of oxygen atoms is \(\frac{5}{2}\). In \(3CO_{2}\), the number of oxygen atoms is \(6\). Let the coefficient of \(O_{2}\) be \(x\).
The total number of oxygen atoms on the right - hand side is \(\frac{5}{2}+6 + 2x\).
We know that \(9=\frac{5}{2}+6 + 2x\).
\(9=\frac{5 + 12}{2}+2x\), \(9=\frac{17}{2}+2x\), \(2x=9-\frac{17}{2}=\frac{18 - 17}{2}=\frac{1}{2}\), \(x=\frac{1}{4}\)
Multiply the entire equation by \(4\) to get rid of the fractions.

Answer:

\(4C_{3}H_{5}(NO_{3})_{3}(l)
ightarrow6N_{2}(g)+O_{2}(g)+10H_{2}O(g)+12CO_{2}(g)\)