A mixture of ammonia and oxygen is prepared by combining 0.250 L of NH3 (measured at 0.700 atm and 20°C) with 0.160 L of O2 (measured at 0.720 atm and 49°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)
I get that oxygen is the limiting factor
For NH3 I get n=.00727
For o2 I get n= .00435
So I do 3/2 x .00435 =.0029
Then .740atm x L = .0029 x .0821 x 373
Then .0888/.740 =.120
Does this look right?
Some picky points.
n N2 I obtained 0.007279 so I rounded to 0.00728.
For n O2 I obtained 0.00436
I obtained 0.0029 for mols N2 just as you did; however, it is done by 0.0036 x 2/3 and not 3/2.
Finally,(this is not so picky) the problem asks for mL so you should convert your 0.120L to 120 mL.
Thank you! It was right!!!
To find the number of milliliters of N2 that could be formed, we need to set up a stoichiometry calculation. Here's how to do it step by step:
Step 1: Calculate the number of moles of NH3 (ammonia):
The given volume of NH3 is 0.250 L, measured at 0.700 atm and 20°C. To convert the given conditions to standard temperature and pressure (STP), we need to use the ideal gas law equation:
PV = nRT
First, convert the temperature to Kelvin: 20°C + 273.15 = 293.15 K
Now, substitute the values into the equation:
(0.700 atm)(0.250 L) = n(0.0821 L·atm/mol·K)(293.15 K)
Solve for n (moles of NH3):
n = (0.700 atm × 0.250 L) / (0.0821 L·atm/mol·K × 293.15 K) = 0.00727 moles
So, you correctly calculated the moles of NH3 as 0.00727.
Step 2: Calculate the number of moles of O2 (oxygen):
The given volume of O2 is 0.160 L, measured at 0.720 atm and 49°C. We'll follow the same approach as in Step 1 to convert the given conditions to STP:
Convert the temperature to Kelvin: 49°C + 273.15 = 322.15 K
Substitute the values into the ideal gas law equation:
(0.720 atm)(0.160 L) = n(0.0821 L·atm/mol·K)(322.15 K)
Solve for n (moles of O2):
n = (0.720 atm × 0.160 L) / (0.0821 L·atm/mol·K × 322.15 K) = 0.00435 moles
So, you correctly calculated the moles of O2 as 0.00435.
Step 3: Determine the limiting reactant:
To determine the limiting reactant, we need to compare the stoichiometric coefficients in the balanced equation with the moles of each reactant.
From the balanced equation 4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(g), we see that the stoichiometric ratio between NH3 and O2 is 4:3.
Comparing the moles of NH3 and O2, we have 0.00727 moles of NH3 and 0.00435 moles of O2. The ratio of NH3 to O2 is:
0.00727 moles NH3 / 0.00435 moles O2 = 1.67
Since the ratio is less than 4:3, NH3 is in excess, and oxygen is the limiting reactant.
Step 4: Calculate the moles of N2 formed:
Using the stoichiometry of the balanced equation, we know that 3 moles of O2 react to form 2 moles of N2. From the previous step, we found that we have 0.00435 moles of O2.
Using this information, we set up a stoichiometry calculation:
(2 moles N2 / 3 moles O2) × 0.00435 moles O2 = 0.00290 moles N2
Thus, correctly calculating the moles of N2 as 0.00290.
Step 5: Convert moles of N2 to milliliters at the given condition:
Given that the volume of N2 is measured at 0.740 atm and 100°C, we need to convert the moles to milliliters using the ideal gas law.
First, convert the temperature to Kelvin: 100°C + 273.15 = 373.15 K
Now, substitute the values into the ideal gas law equation:
(0.740 atm)(V) = (0.00290 moles)(0.0821 L·atm/mol·K)(373.15 K)
Solve for V (volume of N2 in liters):
V = (0.00290 moles × 0.0821 L·atm/mol·K × 373.15 K) / 0.740 atm ≈ 0.122 L
Finally, convert liters to milliliters:
0.122 L × (1000 mL / 1 L) ≈ 122 mL
So, the volume of N2 that could be formed is approximately 122 milliliters.
Thus, your final step seems incorrect. The correct answer is approximately 122 mL, not 0.120.