Answer:
15.8528 grams of oxygen will be obtained.
Explanation:
[tex]4C_3H_5N_3O_9\rightarrow 6N_2+12CO_2+10H_2O+O_2[/tex]
Mass of nitroglycerin = [tex]4.50\times 10^{2} g[/tex]
Moles of nitroglycerin:
[tex]\frac{4.50\times 10^{2} g}{227.08 g/mol}=1.9816 mol[/tex]
According to reaction, 4 moles of nitroglycerin gives 1 mole of oxygen gas.
then 1.9816 mol of nitroglycerin will give:
[tex]\frac{1}{4}\times 1.9816=0.4954 [/tex] moles of oxygen gas.
Mass of oxygen ;
[tex]0.4954 mol\times 32 g/mol=15.8528 g[/tex]
15.8528 grams of oxygen will be obtained.
The maximum mass of O₂ that can be obtained from 4.5×10² g of nitroglycerin is 15.86 g
The balanced equation for the reaction is given below:
Next, we shall determine the mass of C₃H₅N₃O₉ that decomposed and the mass of O₂ produced from the balanced equation. This can be obtained as follow:
Molar mass of C₃H₅N₃O₉ = (12×3) + (5×1) + (14×3) + (16×9)
= 36 + 5 + 42 + 144
= 227 g/mol
Molar mass of O₂ = 2 × 16 = 32 g/mol
From the balanced equation above,
908 g of C₃H₅N₃O₉ decomposed to produce 32 g of O₂.
Finally, we shall determine the mass of O₂ that will be produce from the decomposition of 4.5×10² g of C₃H₅N₃O₉. This can be obtained as follow:
From the balanced equation above,
908 g of C₃H₅N₃O₉ decomposed to produce 32 g of O₂.
Therefore, 4.5×10² g of C₃H₅N₃O₉ will decompose to produce = [tex]\frac{4.5*10^{2} * 32}{908}[/tex] = 15.86 g of O₂
Thus, the maximum mass of O₂ obtained from 4.5×10² g of nitroglycerin is 15.86 g
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