Urea, NH2CONH2, is manufactured from ammonia and carbon dioxide according to the following equation:
2 NH3 (g) + CO2 (g) -> NH2CONH2(aq)+ H2O (l)
What volume of ammonia gas at 25 o C and 1.5 atm is needed to make 500g of urea?
Convert 500 g urea to mols. mols = grams/molar mass.
Convert mols urea to mols NH3 using the coefficients in the balanced equation.
Use PV = nRT to convert mols NH3 at STP (mols you have from step 2) to mols NH3 at 25 C and 1.5 atm.
Post your work if you get stuck.
271L
91L
To find the volume of ammonia gas required to make 500 grams of urea, we can use the ideal gas law equation, which states:
PV = nRT,
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
First, we need to convert the given temperature to Kelvin:
25 degrees Celsius + 273.15 = 298.15 Kelvin.
Next, we need to determine the number of moles of urea using its molar mass. The molar mass of urea (NH2CONH2) is:
(2 * 1.008) + 12.010 + (2 * 14.007) + (4 * 1.008) = 60.055 g/mol.
Therefore, the number of moles of urea is:
500 g / 60.055 g/mol = 8.33 moles.
Now, let's look at the balanced chemical equation:
2 NH3 (g) + CO2 (g) -> NH2CONH2 (aq) + H2O (l)
Based on the equation, we can see that it takes two moles of ammonia gas (NH3) to produce one mole of urea (NH2CONH2).
Therefore, the number of moles of ammonia required is:
8.33 moles * 2 = 16.67 moles.
Finally, we can use the ideal gas law equation to calculate the volume of ammonia gas:
PV = nRT,
where P = 1.5 atm, V is the volume we want to find, n = 16.67 moles, R = 0.0821 L * atm/(mol * K), and T = 298.15 K.
Substituting the values into the equation, we have:
(1.5 atm) * V = (16.67 moles) * (0.0821 L * atm/(mol * K)) * (298.15 K).
V = [(16.67 * 0.0821 * 298.15) / 1.5] L.
V ≈ 678.61 L.
Therefore, approximately 678.61 liters of ammonia gas at 25 oC and 1.5 atm are needed to make 500g of urea.