Answer:
a. 561L of N₂
b. 14,80L of CO₂
c. 715,0L of gases
Explanation:
The explosion of nitroglycerine gives the reaction:
4C₃H₅N₃O₉(s) → 6N₂(g) + 12CO₂(g) + 10H₂O(g) + O₂(g)
a. 4 moles of nitroglycerine produce 6 moles of N₂. Thus, 16,7 moles of nitrofglycerine produce:
16,7 moles C₃H₅N₃O₉ × (6 moles N₂ / 4 moles C₃H₅N₃O₉) = <em><u>25,05 moles N₂</u></em>
At STP, 1 mole of gas occupies 22,4 L. 25,05 moles are:
25,05 moles N₂ × (22,4L / 1mole) = <em>561 L</em>
b. 100,0g of C₃H₅N₃O₉ are:
100,0g C₃H₅N₃O₉ × (1mole / 227,1g) = 0,4403 moles of C₃H₅N₃O₉.
As 4 moles of C₃H₅N₃O₉ produce 6 mole of CO₂. Moles of CO₂ are:
0,4403 moles of C₃H₅N₃O₉ × ( 6mol CO₂ / 4mol C₃H₅N₃O₉) = 0,6605 moles CO₂ Again, as 1 mol of gas occupies 22,4L:
0,6605 mol CO₂ × (22,4L / 1mol) = <em>14,80 L</em>
c.<em> </em>1,000 kg ≈ 1000g of nytroglicerine are:
1000g C₃H₅N₃O₉ × (1mole / 227,1g) = 4,403 moles of C₃H₅N₃O₉.
As 4 moles of C₃H₅N₃O₉ produce 29 moles of gases. Moles of CO₂ are:
4,403 moles of C₃H₅N₃O₉ × ( 29mol gases / 4mol C₃H₅N₃O₉) = 31,92 moles of gases. Again, as 1 mol of gas occupies 22,4L:
31,92 mol CO₂ × (22,4L / 1mol) = <em>715,0 L </em>
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I hope it helps!
