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.
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?
3 answers
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?
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?
4 answers