Ammonia gas and carbon dioxide gas react to form aqueous urea (NH2CoNH2) and water. What volume of ammonia at 25 degrees celcius and 1.5 atm pressure is needed to produce 500 g of urea?
PV = nRT where R = 0.0821 atm.L / K.mol
NH2CONH2 = (14.01)(2)+(1.01)(4)+(12.01)+(16) = 60.07
n = 500 g/60.07 g = 8.32362244049 = 8.3236 mol of NH2CONH2 = 16.6472 mol of NH3
(1.5)V = (16.6472)(0.0821 atm L / K mol)(298.15 K)
(1.5)V = 407.492076028
V = 271.661384019
271.66 L of ammonia gas
Why did the ammonia and carbon dioxide go on a date?
Because together they make a great "aqua-rea"!
Ahem, let's get serious now. To solve this problem, we need to start by balancing the chemical equation:
2 NH3(g) + CO2(g) -> (NH2)2CO(aq)
According to the stoichiometry, we need 2 moles of ammonia for every 1 mole of urea. Now, let's do some calculations.
1. Convert the mass of urea to moles:
Molar mass of urea (NH2CoNH2) = 60.06 g/mol (approximate)
Moles of urea = mass of urea / molar mass
= 500 g / 60.06 g/mol
≈ 8.33 mol
2. Determine the moles of ammonia needed:
From the balanced equation, we know that 2 moles of ammonia yield 1 mole of urea. Therefore:
Moles of ammonia needed = 2 * moles of urea
≈ 2 * 8.33 mol
≈ 16.66 mol
3. Apply the ideal gas law to find the volume of ammonia:
PV = nRT, where P is pressure, V is volume, n is number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is temperature in Kelvin.
Substituting the values:
V = (nRT) / P
= (16.66 mol * 0.0821 L·atm/mol·K * (25 + 273)) / 1.5 atm
≈ 880 L
So, approximately 880 liters of ammonia gas at 25 degrees Celsius and 1.5 atm pressure would be needed to produce 500 g of urea. Keep in mind that this is a simplified calculation and may not account for all factors.
To answer this question, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (K)
First, let's calculate the number of moles of urea produced from 500g. To do that, we need to know the molar mass of urea.
The molar mass of NH2CoNH2 (urea) is:
(2 x 1) + (1 x 12) + (2 x 14) + (2 x 16) = 60 g/mol
Number of moles of urea:
n = mass / molar mass
n = 500 g / 60 g/mol
n = 8.33 mol
Since the reaction is balanced, it means that one mole of urea is produced for every mole of ammonia.
Therefore, we need 8.33 moles of ammonia to produce 8.33 moles of urea.
Now let's calculate the volume of ammonia needed. We can rearrange the ideal gas law equation to solve for volume:
V = nRT / P
V = (8.33 mol) x (0.0821 L·atm/(mol·K)) x (25°C + 273.15 K) / 1.5 atm
V ≈ 365.82 L
Therefore, the volume of ammonia gas needed at 25 degrees Celsius and 1.5 atm pressure to produce 500 g of urea is approximately 365.82 liters.
To determine the volume of ammonia gas needed to produce 500 g of urea, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
First, let's convert the given temperature from degrees Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 25 + 273.15
T(K) = 298.15 K
Next, we need to calculate the number of moles of urea using its molar mass. The molar mass of urea (NH2CONH2) is:
Molar mass (urea) = 14.01 g/mol + 2(1.01 g/mol) + 12.01 g/mol + 16.00 g/mol + 14.01 g/mol + 2(1.01 g/mol)
Molar mass (urea) = 60.06 g/mol
Number of moles (urea) = mass (urea) / molar mass (urea)
Number of moles (urea) = 500 g / 60.06 g/mol
Number of moles (urea) = 8.326 mol
Now, let's look at the balanced chemical equation for the reaction:
CO2 + 2NH3 → NH2CONH2 + H2O
From the equation, we can see that 2 moles of ammonia (NH3) react to produce 1 mole of urea (NH2CONH2).
Using the stoichiometry of the reaction, we can determine the number of moles of ammonia (NH3) required to produce 8.326 mol of urea (NH2CONH2):
Number of moles (ammonia) = 2 * Number of moles (urea)
Number of moles (ammonia) = 2 * 8.326 mol
Number of moles (ammonia) = 16.652 mol
Finally, we can use the ideal gas law equation to calculate the volume of ammonia gas at 25°C and 1.5 atm pressure:
V = (n * R * T) / P
V = (16.652 mol * 0.0821 L·atm/mol·K * 298.15 K) / 1.5 atm
V ≈ 670.68 L
Therefore, approximately 670.68 liters of ammonia gas at 25 degrees Celsius and 1.5 atm pressure are needed to produce 500 g of urea.