Ammonia is not the only possible fertilizer. Others include urea, which can be produced by the reaction, CO2(g) + 2 NH3(g) --> CO(NH2)2 (s) + H2O(g).
A scientist has 75g of dry ice to provide the carbon dioxide. If 4.50L of ammonia at 15C and a pressure of 1.4atm is added, which reactant is limiting? What mass of urea will form?
To determine which reactant is limiting, we need to compare the amount of moles of each reactant and their stoichiometric coefficients in the balanced equation.
First, let's calculate the moles of dry ice (CO2) using its molar mass:
Molar mass of CO2 = 12.01 g/mol + 2(16.00 g/mol) = 44.01 g/mol
Moles of CO2 = mass of dry ice / molar mass of CO2
Moles of CO2 = 75 g / 44.01 g/mol
Next, let's calculate the moles of ammonia (NH3) using the ideal gas law:
PV = nRT
R is the ideal gas constant, which is 0.0821 L·atm/(mol·K)
T is the temperature in Kelvin (15°C + 273.15)
n = (PV) / (RT)
n = (1.4 atm) * (4.50 L) / (0.0821 L·atm/(mol·K) * (15°C + 273.15))
Now, we need to compare the moles of CO2 and NH3 to determine the limiting reagent.
In the balanced equation: CO2(g) + 2NH3(g) → CO(NH2)2(s) + H2O(g)
From the balanced equation, we can see that 1 mole of CO2 reacts with 2 moles of NH3 to produce 1 mole of CO(NH2)2. Therefore, the mole ratio of CO2 to NH3 is 1:2.
If the moles of CO2 are less than half the moles of NH3, then CO2 is the limiting reagent. Otherwise, NH3 is the limiting reagent.
Finally, once we determine the limiting reagent, we can calculate the mass of urea formed.
To calculate the mass of urea (CO(NH2)2), we need to use the molar mass of urea:
Molar mass of CO(NH2)2 = 12.01 g/mol + 2(14.01 g/mol) + 2(1.01 g/mol)
Mass of urea (CO(NH2)2) = moles of limiting reagent × molar mass of urea
Please provide the moles of CO2 and NH3, and I can help you calculate the limiting reagent and the mass of urea formed.
To determine which reactant is limiting in the given reaction, we need to compare the number of moles of carbon dioxide produced from the dry ice and the number of moles of ammonia available.
Step 1: Calculate the moles of carbon dioxide (CO2) from the given mass of dry ice.
The molar mass of CO2 is 44.01 g/mol.
Moles of CO2 = mass of dry ice (g) / molar mass of CO2 (g/mol)
Moles of CO2 = 75 g / 44.01 g/mol
Step 2: Calculate the moles of ammonia from the given conditions.
We need to convert the given volume of ammonia at 15°C and 1.4 atm to moles using the ideal gas law.
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.
1 atm = 101.325 kPa = 101325 Pa
Convert 15°C to Kelvin: T(K) = 15 + 273.15
n = PV / RT
n = (1.4 atm * 101325 Pa/atm) / (R * (15 + 273.15) K)
Step 3: Compare the number of moles of CO2 and NH3 to find the limiting reactant.
Since the balanced equation shows that 1 mole of CO2 requires 2 moles of NH3, we need to compare the moles of CO2 with half the moles of NH3.
Multiply the moles of ammonia by 0.5 to get half the moles:
Moles of NH3 (half) = 0.5 * moles of NH3
If the moles of CO2 is less than moles of NH3 (half), then CO2 is the limiting reactant. Otherwise, NH3 is the limiting reactant.
Step 4: Calculate the mass of urea formed using the limiting reactant.
The molar mass of urea (CO(NH2)2) is 60.06 g/mol.
The balanced equation states that 1 mole of CO2 reacts to produce 1 mole of urea.
Mass of urea = moles of CO2 * molar mass of urea (g/mol)
Following these steps, you can determine which reactant is limiting and calculate the mass of urea that will form.
where did 22.4 L come from? and how do I get the mass?
CO2(g) + 2 NH3(g) --> CO(NH2)2 (s) + H2O(g).
mols CO2 = grams/molar mass = ?
mols NH3 = 4.50 L x (1 mol/22.4L) = ?
Convert mols CO2 to mols urea produced = 1:1 or mols NH3 = mols urea.
Convert mols NH3 to mols urea produced = 2 mol NH3 = 1 mol urea
In limiting reagent problems (LR) the mols product formed is the SMALLER of the two possibilities. The eagent producing the smaller number of mols or product is the LR.
Post your work if you get stuck.