How many grams of oxygen gas can be produced when 4.50 L of NO2 react according to the equation below at 427 ¨¬C and 1.15 atm?
4NO2(g) + 6H2O(g) ¡æ 7 O2(g) + 4NH3(g)
first figure the moles of NO2 (use the ideal gas law).
then you get 7/4 moles O2 for each mole of NO2. Change that to grams O2
To find the number of grams of oxygen gas produced, we need to use the Ideal Gas Law to calculate the number of moles of NO2 and then use the stoichiometry of the balanced equation to find the moles of O2 produced.
Step 1: Convert liters to moles of NO2
We can use the Ideal Gas Law equation: PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L•atm/mol•K)
T = temperature (in Kelvin)
First, convert the given temperature from degrees Celsius to Kelvin:
T = 427 ºC + 273.15 = 700.15 K
Now we can plug in the values into the Ideal Gas Law equation:
n(NO2) = (P * V) / (R * T)
n(NO2) = (1.15 atm * 4.50 L) / (0.0821 L•atm/mol•K * 700.15 K)
n(NO2) ≈ 0.079 mol
Step 2: Use stoichiometry to find moles of O2
According to the balanced equation, the mole ratio between NO2 and O2 is 4:7. This means that for every 4 moles of NO2, 7 moles of O2 are produced.
Using the mole ratio, we can calculate the moles of O2 produced:
n(O2) = (7/4) * n(NO2)
n(O2) = (7/4) * 0.079 mol
n(O2) ≈ 0.138 mol
Step 3: Convert moles to grams
To convert moles of O2 to grams, we need to use the molar mass of oxygen gas (O2), which is approximately 32 g/mol.
Mass(O2) = n(O2) * molar mass(O2)
Mass(O2) = 0.138 mol * 32 g/mol
Mass(O2) ≈ 4.4 g
Therefore, approximately 4.4 grams of oxygen gas can be produced when 4.50 L of NO2 react according to the given equation at 427 ºC and 1.15 atm.
To find the number of grams of oxygen gas (O2) produced when 4.50 L of NO2 reacts, you will need to use the Ideal Gas Law equation:
PV = nRT
where:
P is the pressure (in atm)
V is the volume (in liters)
n is the number of moles
R is the ideal gas constant (0.0821 L·atm/mol·K)
T is the temperature (in Kelvin)
First, convert the given temperature of 427 ¨¬C to Kelvin:
T (K) = 427 ¨¬C + 273.15 = 700.15 K
Next, rearrange the Ideal Gas Law equation to solve for n (moles):
n = PV / (RT)
Substitute the given values into the equation:
P = 1.15 atm
V = 4.50 L
R = 0.0821 L·atm/mol·K
T = 700.15 K
n = (1.15 atm * 4.50 L) / ((0.0821 L·atm/mol·K) * 700.15 K)
Next, you need to determine the stoichiometry of the given reaction in order to find the moles of oxygen gas produced. From the balanced equation:
4NO2(g) + 6H2O(g) ¡æ 7 O2(g) + 4NH3(g)
The stoichiometric ratio between NO2 and O2 is 4:7. This means that for every 4 moles of NO2, 7 moles of O2 are produced.
Multiply the moles of NO2 (determined above) by the stoichiometric ratio to find the moles of O2:
moles of O2 = moles of NO2 * (7 moles O2 / 4 moles NO2)
Finally, to find the grams of O2 produced, you need to multiply the moles of O2 by its molar mass, which is approximately 32 g/mol.
grams of O2 = moles of O2 * molar mass of O2
Now you have the complete process for finding the number of grams of oxygen gas produced when 4.50 L of NO2 reacts at 427 ¨¬C and 1.15 atm.