Question

nitroglycerin is a dangerous powerful explosive that violently decomposes when it is shaken or dropped. the swedish chemist alfred nobel (1833 - 1896) founded the nobel prizes with a fortune he made by inventing dynamite, a mixture of nitroglycerin and inert ingredients that was safe to handle.

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.
2. suppose 65.0 l of carbon dioxide gas are produced by this reaction, at a temperature of $-13.0^{circ}c$ and pressure of exactly 1 atm. calculate the mass of nitroglycerin that must have reacted. be sure your answer has the correct number of significant digits.

g

Explanation:

1. Balanced Chemical Equation

Step1: Write the unbalanced equation

The decomposition of liquid nitroglycerin \(\ce{C3H5(NO3)3(l)}\) into gaseous dinitrogen \(\ce{N2(g)}\), gaseous dioxygen \(\ce{O2(g)}\), gaseous water \(\ce{H2O(g)}\) and gaseous carbon dioxide \(\ce{CO2(g)}\). The unbalanced equation is:
\(\ce{C3H5(NO3)3(l)
ightarrow N2(g) + O2(g)+H2O(g)+CO2(g)}\)

Step2: Balance the equation

• Balance \(C\) atoms: There are \(3\) \(C\) atoms in \(\ce{C3H5(NO3)3}\), so put a coefficient of \(3\) in front of \(\ce{CO2}\).
• Balance \(N\) atoms: There are \(3\) \(N\) atoms in \(\ce{C3H5(NO3)3}\), so put a coefficient of \(\frac{3}{2}\) in front of \(\ce{N2}\).
• Balance \(H\) atoms: There are \(5\) \(H\) atoms in \(\ce{C3H5(NO3)3}\), so put a coefficient of \(\frac{5}{2}\) in front of \(\ce{H2O}\).
• Now, for \(O\) atoms:

Left - hand side: \(9\) \(O\) atoms from \(\ce{NO3}\) groups.
Right - hand side: \(3\times2\) (from \(\ce{CO2}\))+\(\frac{5}{2}\times1\) (from \(\ce{H2O}\))+\(2x\) (from \(\ce{O2}\))+\(\frac{3}{2}\times0\) (from \(\ce{N2}\)).
Multiply the entire equation by \(4\) to get rid of fractions.
The balanced equation is:
\(\ce{4C3H5(NO3)3(l)
ightarrow 6N2(g) + O2(g)+10H2O(g)+12CO2(g)}\)

Step1: Convert temperature to Kelvin

Use the formula \(T = t+273.15\). Given \(t=- 13.0^{\circ}C\), then \(T=( - 13.0 + 273.15)K=260.15K\)

Step2: Use the ideal gas law \(PV = nRT\) to find the number of moles of \(\ce{CO2}\)

We know that \(P = 1\space atm\), \(V = 65.0L\), \(R=0.0821\space L\cdot atm/(mol\cdot K)\), \(T = 260.15K\)
From \(n=\frac{PV}{RT}\), substitute the values:
\(n=\frac{1\space atm\times65.0L}{0.0821\space L\cdot atm/(mol\cdot K)\times260.15K}\)
\(n=\frac{65.0}{0.0821\times260.15}\space mol\)
\(n=\frac{65.0}{21.36}\space mol\approx3.04\space mol\)

Step3: Use the mole ratio from the balanced equation

From the balanced equation \(\ce{4C3H5(NO3)3(l)
ightarrow 12CO2(g)}\), the mole ratio \(n(\ce{C3H5(NO3)3}):n(\ce{CO2})=\frac{4}{12}=\frac{1}{3}\)
If \(n(\ce{CO2}) = 3.04\space mol\), then \(n(\ce{C3H5(NO3)3})=\frac{3.04}{3}\space mol\approx1.01\space mol\)

Step4: Calculate the molar mass of \(\ce{C3H5(NO3)3}\)

The molar mass of \(\ce{C3H5(NO3)3}\):
\(M=(3\times12.01 + 5\times1.01+3\times14.01 + 9\times16.00)g/mol\)
\(M=(36.03+5.05 + 42.03+144.00)g/mol=227.11g/mol\)

Step5: Calculate the mass of \(\ce{C3H5(NO3)3}\)

Use \(m = nM\), where \(n = 1.01\space mol\) and \(M = 227.11g/mol\)
\(m=1.01\space mol\times227.11g/mol\approx229g\)

Answer:

\(\ce{4C3H5(NO3)3(l)
ightarrow 6N2(g) + O2(g)+10H2O(g)+12CO2(g)}\)

2. Mass of Nitroglycerin