A mixture of ammonia and oxygen is prepared by combining 0.330 L of NH3 (measured at 0.790 atm and 22°C) with 0.210 L of O2 (measured at 0.820 atm and 48°C). How many milliliters of N2 (measured at 0.740 atm and 100. °C) could be formed if the following reaction occurs?

Question

A mixture of ammonia and oxygen is prepared by combining 0.330 L of NH3 (measured at 0.790 atm and 22°C) with 0.210 L of O2 (measured at 0.820 atm and 48°C). How many milliliters of N2 (measured at 0.740 atm and 100. °C) could be formed if the following reaction occurs?

4NH3(g) + 3O2(g) -> 2N2(g) + 6H2O(g)

Answer 1

Use PV =nRT and convert NH3 and O2 (each) to mols = n.

Find the limiting regent
Convert to mols N2 formed, the use PV = nRT and convert to mL at the conditions listed.

Answer 2

To answer this question, we need to use the ideal gas law and stoichiometry. Let's break it down step by step.

Step 1: Convert temperatures to Kelvin.
Given temperatures: 22 °C and 48 °C
To convert to Kelvin, add 273 to each temperature:
22 °C + 273 = 295 K
48 °C + 273 = 321 K

Step 2: Convert pressures to atmospheres.
Given pressures: 0.790 atm and 0.820 atm

Step 3: Calculate the number of moles of NH3 and O2.
To do this, we can use the ideal gas law equation: PV = nRT
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin.

For NH3:
P = 0.790 atm
V = 0.330 L
T = 295 K
R = 0.0821 L·atm/(mol·K)

Rearranging the equation, we have:
n = PV / RT

Plugging in the values:
n = (0.790 atm)(0.330 L) / (0.0821 L·atm/(mol·K))(295 K)
n = 0.00973 mol

For O2, using the same equation:
P = 0.820 atm
V = 0.210 L
T = 321 K
R = 0.0821 L·atm/(mol·K)

n = (0.820 atm)(0.210 L) / (0.0821 L·atm/(mol·K))(321 K)
n = 0.00713 mol

So, we have 0.00973 mol of NH3 and 0.00713 mol of O2.

Step 4: Use the stoichiometry of the balanced equation to find the number of moles of N2 produced.
From the balanced equation, we can see that the mole ratio between NH3 and N2 is 4:2, or 2:1.
Since there are 0.00973 mol of NH3, the number of moles of N2 produced will be half that amount:
0.00973 mol / 2 = 0.00487 mol of N2

Step 5: Convert the number of moles of N2 to volume using the ideal gas law.
To do this, we'll again use the ideal gas law equation: PV = nRT

P = 0.740 atm
V = ?
T = 373 K (since it's given at 100 °C, which is equal to 373 K)
R = 0.0821 L·atm/(mol·K)
n = 0.00487 mol

Rearranging the equation, we have:
V = nRT / P

Plugging in the values:
V = (0.00487 mol)(0.0821 L·atm/(mol·K))(373 K) / (0.740 atm)
V ≈ 0.192 L ≈ 192 mL

So, approximately 192 mL of N2 could be formed.