I have almost the same question as Josh from August 23rd.
If you use 10.0 g of CuSO4 and excess NH3, what is the theoretical yield of Cu(NH3)4SO4?
I can handle the theoretical yield, but how do I find the mass of NH3 so I can find the thoretical yield?
I am confused by your statement. You don't need the mass of NH3. The problem states an excess of NH3 was used; therefore, you know there was sufficient NH3 present for the reaction of all 10 g CuSO4.
CuSO4 + 4NH3 ==> Cu(NH3)4SO4
1. Convert 10 g CuSO4 to mols CuSO4. Remember mols = g/molar mass.
2. Using the equation, convert mols CuSO4 to mols Cu(NH3)4SO4.
3. Convert mols Cu(NH3)4SO4 to g by
mols x molar mass = grams. That will give you the theoretical yield of Cu(NH3)4SO4.
1.365456534
The deep blue compound Cu(NH3)4SO4 is made by the reaction of copper(II) sulfate and ammonia. CuSO4(aq) + 4 NH3(aq) → Cu(NH3)4SO4(aq)
a. If you use 10.0 g of CuSO4 and excess NH3, what is the theoretical yield of Cu(NH3)4SO4?
b. If you isolate 12.6 g of Cu(NH3)4SO4, what is the percent yield of Cu(NH3)4SO4?
To find the mass of NH3 required for the reaction, you need to determine the stoichiometry of the reaction between CuSO4 and NH3. The balanced equation for the reaction is:
CuSO4 + 4NH3 → Cu(NH3)4SO4
From the equation, you can see that 1 mole of CuSO4 reacts with 4 moles of NH3.
Step 1: Convert the mass of CuSO4 provided (10.0 g) to moles:
Molar mass of CuSO4 = atomic mass of Cu + atomic mass of S + 4 * atomic mass of O
= (63.55 g/mol + 32.07 g/mol + 4 * 16.00 g/mol)
= 159.63 g/mol
Moles of CuSO4 = Mass of CuSO4 / Molar mass of CuSO4
= (10.0 g) / (159.63 g/mol)
= 0.0627 mol
Step 2: Use the stoichiometric ratio to determine moles of NH3:
From the balanced equation, the stoichiometric ratio between CuSO4 and NH3 is 1:4.
Moles of NH3 = Moles of CuSO4 * (4 moles of NH3 / 1 mole of CuSO4)
= (0.0627 mol) * (4)
= 0.251 mol
Step 3: Convert moles of NH3 to mass:
Molar mass of NH3 = atomic mass of N + 3 * atomic mass of H
= (14.01 g/mol + 3 * 1.01 g/mol)
= 17.03 g/mol
Mass of NH3 = Moles of NH3 * Molar mass of NH3
= (0.251 mol) * (17.03 g/mol)
= 4.26 g
So, the mass of NH3 required for the reaction is approximately 4.26 grams.