Answer : The maximum amount of [tex]O_{2}[/tex] obtained = 12.576 g
Solution : Given,
Mass of Nitroglycerin [tex]C_{3} H_{5} N_{3} O_{9}[/tex] = 3.50 × 102g
Molar mass of Nitroglycerin [tex]C_{3} H_{5} N_{3} O_{9}[/tex] = 227.0865 g/mol
First, find the number of moles of Nitroglycerin i.e
Number of moles of Nitroglycerin = [tex]\frac{Given Mass}{Molar Mass}[/tex]
= [tex]\frac{3.50\times 102g}{227.0865g/mol}[/tex]
= 1.5720 moles
[tex]4C_{3} H_{5} N_{3} O_{9}\rightarrow 6N_{2}+12Co_{2}+10H_{2}O+O_{2}[/tex]
According to reaction,
4 moles of Nitroglycerin gives 1 mole of Oxygen
and, 1.5720 moles of Nitroglycerin gives → [tex]\frac{1}{4}\times 1.5720[/tex]
= 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