Chapter 13, Problem 103A

Analyze When nitroglycerin $(C_3{H}_5{N}_3{O}_9)$ explodes, itdecomposes into the following gases: $C{O}_2$ , $N_2$ , NO, and $H_{2}O$ . If 239 g of nitroglycerin explodes, what volumewill the mixture of gaseous products occupy at 1.00 atmpressure and 2678°C?

Interpretation Introduction

Interpretation:

The volume of mixture of products produced by $239g$ of nitroglycerin is to be calculated.

Concept introduction:

Stochiometry concepts are used to calculate the result.

Answer to Problem 103A

$1850L$ of mixture of product is formed.

Explanation of Solution

The given question involves the chemical equation as:

$4C_3{H}_5{N}_3{O}_9\to 5N_2+2NO+10H_{2}O+12C{O}_2$

Molecular mass of nitroglycerin = $227g/mol$

According to Avogadro’s principle,

$\begin{array}{c}numberofmolesofnitroglycerin=\frac{239}{227}\\ =1.05mol\end{array}$

Total gas products=29

$\begin{array}{c}numberofmolesofgasproducts=\frac{29}{4}\times 1.05\\ =7.61mol\end{array}$

Now, volume is given by:

$\begin{array}{l}V=\frac{nRT}{P}\\ P=1atm\\ T=2678°C\\ =2678+273\\ =2951K\\ R=0.0821Latm{K}^{-1}mo{l}^{-1}\end{array}$

Applying the formula:

$\begin{array}{l}V=\frac{nRT}{P}\\ =\frac{7.61\times 0.0821\times 2951}{1}\\ =1850L\end{array}$

Conclusion

$1850L$ of mixture of product is formed.