How many moles of nitroglycerin were in the container originally

Asked by admin @ 02/12/2021 in Chemistry viewed by 329 People

When nitroglycerine (227.1 g/mol) explodes, N2, CO2, H2O, and O2 gases are released initially. Assume that the gases from the explosion cool to standard conditions without reacting further. 4 C3H5N3O9 (s) yields 6 N2 (g) + 12 CO2 (g) + 10 H2O (g) + O2 (g). a. If 16.7 moles of nitroglycerine react, how many liters of N2 are produced? b. What volume of CO2 (at STP) will be produced when 100.0 g of nitroglycerine reacts? c. What is the total volume of gas (at STP) produced when 1.000 kg of nitroglycerine reacts? Please show work.

Answered by admin @ 02/12/2021

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) = 25,05 moles N₂

At STP, 1 mole of gas occupies 22,4 L. 25,05 moles are:

25,05 moles N₂ × (22,4L / 1mole) = 561 L

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) = 14,80 L

c. 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) = 715,0 L

I hope it helps!

How many moles of nitroglycerin were in the container originally

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