When nitroglycerin (C3H5N3O9) explodes, it decomposes into the following gases: CO2, N2, NO, and H2O. If 239g of nitroglycerin explodes, what volume will the mixture of gaseous produces occupy at 1.00 atm pressure and 2678oC?

a. Write a balanced equation for the reaction.

b. Use stoichiometry to find the moles of ALL the gas products combined.

c. Use the ideal gas law to solve the volume of the products of the explosion.

How much of this can you do?

A seems easy enough. What do you have for that?
B. What's the mol count?
C. What's the ideal gas law and what do you plug into it?

A. 4C3H5N3O9>12CO2+5N2+2NO+10H2O

Not sure about the other yet. I am trying to figure it out , so I can explain it to my daughter. Is A correct?

a. To write a balanced equation for the reaction, we need to consider the decomposition of nitroglycerin (C3H5N3O9) into the indicated gases. The balanced equation is:

4 C3H5N3O9(s) ---> 6 CO2(g) + 10 H2O(g) + 2 N2(g) + O2(g)

b. To find the moles of all the gas products combined, we need to use stoichiometry. First, we calculate the molar mass of nitroglycerin:

Molar mass of nitroglycerin (C3H5N3O9):
= (3 * 12.01 g/mol) + (5 * 1.01 g/mol) + (3 * 14.01 g/mol) + (9 * 16.00 g/mol)
= 227.08 g/mol

Next, we calculate the moles of nitroglycerin using its given mass:

Moles of nitroglycerin = mass / molar mass
Moles of nitroglycerin = 239 g / 227.08 g/mol
Moles of nitroglycerin ≈ 1.053 mol

From the balanced equation, we can see that for every 4 moles of nitroglycerin that decompose, we produce 6 moles of CO2, 10 moles of H2O, 2 moles of N2, and O2. Hence, the total moles of gas produced will be:

Total moles of gas produced = (6 moles CO2) + (10 moles H2O) + (2 moles N2) + (1 mole O2)
Total moles of gas produced = 6 moles + 10 moles + 2 moles + 1 mole
Total moles of gas produced = 19 moles

c. To solve for the volume of the products of the explosion, we use the ideal gas law equation:

PV = nRT

Where:
P = Pressure (1.00 atm)
V = Volume (which is what we need to find)
n = number of moles of gas (19 moles)
R = Ideal gas constant (0.0821 L∙atm/mol∙K)
T = Temperature (2678°C, which needs to be converted to Kelvin)

The temperature needs to be converted to Kelvin by adding 273.15:

T = 2678°C + 273.15 = 2951.15 K

Now, we can calculate the volume using the ideal gas law equation:

V = nRT / P
V = (19 moles * 0.0821 L∙atm/mol∙K * 2951.15 K) / 1.00 atm
V ≈ 473.08 L

Therefore, the volume of the gaseous products of the explosion will occupy approximately 473.08 liters at 1.00 atm pressure and 2678°C.