4 NH4NO3 -> 8 H2O + 4 N2 + 2 O2

What will be the total volume in liters of gas produced at 720. mm Hg and 570.0°C when 2.50 g of NH4NO3 completely decomposes? Use molar masses with at least as many significant figures as the data given.

Here is an example of a stoichiometry problem. Just follow the steps. You will need to do it three times, convert NH4NO3 each time to H2O, N2, and O2 and get a total number of moles. Then convert moles, using PV = nRT at the conditions listed, to volume. Here is the link.

http://www.jiskha.com/science/chemistry/stoichiometry.html

There is a shorter way. You CAN just do the calculation one time and use the moles from the first calculation to determine all of the others.

The easiest approach is to assume that all the gases are ideal gases, so that 4 moles of NH4NO3 produces 14 moles of gas.

so 2.5 g of NH4NO3 is 2.50/80.0 moles = 0.03125 moles

which will produce 0.03125 x 14/4 moles of gas = 0.1094 moles

which is 22.4 L x0.1094 = 2.45 L at STP

then use P1V1/T1 = P2V2/T2 to find the new total volume.

Alternatively use PV=nRT, where n=0.1094.

To find the total volume of gas produced when 2.50 g of NH4NO3 decomposes, you need to use the ideal gas law equation, PV = nRT. However, since the equation you provided does not mention any volume of gas produced per mole of NH4NO3, we need to first determine the balanced chemical equation for the decomposition process.

The balanced equation you provided is:
4 NH4NO3 -> 8 H2O + 4 N2 + 2 O2

From the equation, we can see that:
- For every 1 mole of NH4NO3, 4 moles of N2 gas are produced.
- For every 1 mole of NH4NO3, 2 moles of O2 gas are produced.

First, let's calculate the number of moles of NH4NO3 in 2.50 g of NH4NO3.

1. Determine the molar mass of NH4NO3:
N = 14.01 g/mol
H = 1.01 g/mol
O = 16.00 g/mol

Molar mass of NH4NO3 = (4 x 1.01) + (1 x 14.01) + (3 x 16.00) = 80.05 g/mol

2. Calculate the number of moles:
moles of NH4NO3 = mass / molar mass = 2.50 g / 80.05 g/mol ≈ 0.031 moles

Now that we know the number of moles of NH4NO3, we can determine the number of moles of gas produced.

- For every 1 mole of NH4NO3, 4 moles of N2 gas are produced:
moles of N2 = 4 x moles of NH4NO3 = 4 x 0.031 moles = 0.124 moles

- For every 1 mole of NH4NO3, 2 moles of O2 gas are produced:
moles of O2 = 2 x moles of NH4NO3 = 2 x 0.031 moles = 0.062 moles

Now, to find the total volume of gas produced, we can use the ideal gas law equation: PV = nRT.

Given:
Pressure (P) = 720. mm Hg
Temperature (T) = 570.0°C = 570.0 + 273.15 = 843.15 K (convert to Kelvin)
R (ideal gas constant) = 0.0821 L·atm/(mol·K) (using consistent units)

Since we have different moles of gases produced, we need to calculate the volume for each gas separately and then sum them up to get the total volume.

1. For N2 gas:
PV = nRT
V = (nRT) / P
V = (0.124 moles x 0.0821 L·atm/(mol·K) x 843.15 K) / (720. mm Hg x 1 atm/760. mm Hg)
V ≈ 0.138 L

2. For O2 gas:
PV = nRT
V = (nRT) / P
V = (0.062 moles x 0.0821 L·atm/(mol·K) x 843.15 K) / (720. mm Hg x 1 atm/760. mm Hg)
V ≈ 0.069 L

Finally, to find the total volume of gas produced, we sum the volumes of N2 and O2:
Total volume = 0.138 L + 0.069 L = 0.207 L

Therefore, the total volume of gas produced at 720. mm Hg and 570.0°C when 2.50 g of NH4NO3 completely decomposes is approximately 0.207 liters.