Urea, which has the chemical formula (NH2)2CO, is a fertilizer that can be prepared by reacting ammonia (NH3) with carbon dioxide (CO2). Given the following chemical equation, what is the theoretical yield of urea (in grams) if 4.16 mol ammonia is the limiting reactant?
2 NH3(g) + CO2(g) → (NH2)2CO(aq) + H2O(l)
To find the theoretical yield of urea, we need to follow these steps:
Step 1: Identify the stoichiometry of the reaction.
From the balanced chemical equation, we can see that it takes a 2:1 ratio of ammonia to urea. This means that for every 2 moles of ammonia, we get 1 mole of urea.
Step 2: Calculate the moles of urea produced.
Since we have 4.16 moles of ammonia, we can use the stoichiometry to determine the moles of urea produced:
4.16 mol NH3 * (1 mol urea / 2 mol NH3) = 2.08 mol urea
Step 3: Convert moles of urea to grams.
To convert from moles to grams, we need to know the molar mass of urea, which is [(14.01 g/mol) * 2 (N) + 12.01 g/mol (C) + 16.00 g/mol + 2 (O)]:
(14.01 * 2) + 12.01 + (16.00 * 2) = 60.06 g/mol
Now we can calculate the theoretical yield:
2.08 mol urea * 60.06 g/mol = 124.9928 g
Therefore, the theoretical yield of urea is approximately 125 grams.
To find the theoretical yield of urea (in grams), we need to follow these steps:
1. Determine the molar ratio between ammonia (NH3) and urea ((NH2)2CO) using the balanced chemical equation:
2 NH3(g) + CO2(g) → (NH2)2CO(aq) + H2O(l)
From the equation, we can see that the molar ratio between NH3 and (NH2)2CO is 2:1.
2. Calculate the number of moles of urea that can be produced using the given amount of ammonia. We are given that the limiting reactant is ammonia, and we have 4.16 mol of ammonia.
Since the molar ratio between NH3 and (NH2)2CO is 2:1, we can say that for every 2 mol of ammonia, we get 1 mol of urea.
Thus, using the given 4.16 mol of ammonia, we can calculate the moles of urea:
4.16 mol NH3 * (1 mol (NH2)2CO / 2 mol NH3) = 2.08 mol (NH2)2CO
3. Convert mol of urea to grams. The molar mass of urea ((NH2)2CO) is:
1 nitrogen atom (N) = 14.01 g/mol
2 hydrogen atoms (H) = 2.02 g/mol (each)
1 carbon atom (C) = 12.01 g/mol
3 oxygen atoms (O) = 16.00 g/mol (each)
So, the molar mass of urea is:
14.01 g/mol + 2 * 1.02 g/mol + 12.01 g/mol + 3 * 16.00 g/mol = 60.06 g/mol
Now, we can calculate the mass of urea:
2.08 mol (NH2)2CO * 60.06 g/mol = 124.89 g (rounded to two decimal places)
Therefore, the theoretical yield of urea is approximately 124.89 grams.
mols NH3 = grams NH3/molar mass NH3 = ?
mols urea = mols NH3 x (1 mol urea/2 mols NH3) = mols NH3 x 1/2 = ?
Then grams urea = mols urea x molar mass urea = ?