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? show work.

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!

Los números delante de las fórmulas son llamados coeficientes estequiométricos. Estos deben ser tales para que la ecuación química esté balanceada, es decir, que el número de átomos de cada elemento de las sustancias reaccionantes y de los productos de la reacción sea el mismo. :))