Asked by Josh Porter
If 8.25 g of Calcium metal reacts with a solution containing 12.50 g of Cu(NO3)2 to produce copper metal, which reactant is limiting? What mass of copper is produced?
Answers
Answered by
DrBob222
Ca(s) + Cu(NO3)2(aq) ==> Cu(s) + Ca(NO3)2(aq)
mols Ca = 8.25/40 = approx 0.21 but you need to do it more accurately.
mol Cu(NO3)2 = 12.5/187 = approx 0.067
You have approx 0.21 mols Ca so you will need approx 0.21 mols Cu(NO3)2. You don't have that much so Cu(NO3)2 must be the limiting reagent.
0.067 mols Cu(NO3)2 will produce 0.067 mols Cu metal.
Then grams Cu metal = mols Cu x atomic mass Cu metal = ? g
mols Ca = 8.25/40 = approx 0.21 but you need to do it more accurately.
mol Cu(NO3)2 = 12.5/187 = approx 0.067
You have approx 0.21 mols Ca so you will need approx 0.21 mols Cu(NO3)2. You don't have that much so Cu(NO3)2 must be the limiting reagent.
0.067 mols Cu(NO3)2 will produce 0.067 mols Cu metal.
Then grams Cu metal = mols Cu x atomic mass Cu metal = ? g
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