How do we represent the decompositon of C_3H_5(NO_3)_3? And if 25.0*g carbon dioxide were collected, what was the starting mass of the nitrate?

Feb 21, 2017

A little too MANY grams for my liking..........

Explanation:

$4 {C}_{3} {H}_{5} {\left(N {O}_{3}\right)}_{3} \left(l\right) \rightarrow 12 C {O}_{2} \left(g\right) + 6 {N}_{2} \left(g\right) + 10 {H}_{2} O \left(g\right) + {O}_{2} \left(g\right)$

Is this balanced? How do you know? How is energy transferred during this reaction?

$\text{Moles of carbon dioxide} = \frac{25.0 \cdot g}{44.0 \cdot g \cdot m o {l}^{-} 1} \cong 0.5 \cdot m o l$

Given the stoichiometry, if there $0.5 \cdot m o l$ carbon dioxide are produced, then $0.167 \cdot m o l$ of the nitrate were consumed.

This represents a mass of,

$0.167 \cdot m o l \times 227.09 \cdot g \cdot m o {l}^{-} 1 \cong 38 \cdot g .$

Funnily enuff, now the major use for this substance is probably as a medication in heart stimulants, and not for munitions and explosives. It's not chemistry that I'd like to do personally.