Question

Be sure to answer all parts. Nitroglycerin (c3h5n3o9) is a powerful explosive. Its decomposition may be represented by 4c3h5n3o9 → 6n2 12co2 10h2o o2 this reaction generates a large amount of heat and gaseous products. It is the sudden formation of these gases, together with their rapid expansion, that produces the explosion. (a) what is the maximum amount of o2 in grams that can be obtained from 3.50 × 102 g of nitroglycerin

Answer 1

Answer : The maximum amount of
`O_(2)` obtained = 12.576 g

Solution : Given,

Mass of Nitroglycerin
`C_(3) H_(5) N_(3) O_(9)` = 3.50 × 102g

Molar mass of Nitroglycerin
`C_(3) H_(5) N_(3) O_(9)` = 227.0865 g/mol

First, find the number of moles of Nitroglycerin i.e

Number of moles of Nitroglycerin =
`(Given Mass)/(Molar Mass)`

=
`(3.50* 102g)/(227.0865g/mol)`

= 1.5720 moles

`4C_(3) H_(5) N_(3) O_(9)\rightarrow 6N_(2)+12Co_(2)+10H_(2)O+O_(2)`

According to reaction,

4 moles of Nitroglycerin gives 1 mole of Oxygen

and, 1.5720 moles of Nitroglycerin gives →
`(1)/(4)* 1.5720`

= 0.393 moles

Now, we have to find the amount of Oxygen obtained.

Formula used :

Weight of Oxygen obtained = Number of moles of Oxygen × Molar mass of oxygen

Weight of Oxygen obtained = 0.393 moles × 32 g/mol

= 12.576 g