This is a limiting reagent (LR) problem and I work these the long way and probably not at all like your prof does. But here goes.
mols Ca = 6.87/atomic mass Ca = approx 0.171 which you have.
mols O2 = 4.58/32 = approx 0.143 ( you used the atomic mass of O and not the molar mass of O2).
The next step is to convert mols Ca to mols CaO AND convert mols O2 to mols CaO.
First Ca to CaO is 0.171 x (2 mol CaO/2 mol CaO) = 0.171 mols CaO formed IF we had all of the O2 we needed.
Next O2 to CaO is 0.143 x (2 mols CaO/1 mol O2) = 0.286 mols CaO IF we had all of the Ca we needed.
These two values are the same so one of them must be wrong. The correct value in LR problems is ALWAYS the smaller value so the correct value is we form 0.171 mols CaO and Ca is the limiting reagent.
Now use the smaller value and grams CaO = mols x molar mass.
Can someone please help me.....
I have limited reactant formula problem. I am completely lost.
2Ca(s) + O2(g) ----> 2CaO(s)
Calculate the mass of Calcium Oxide that can be prepared from 6.87g of Ca and 4.58g of O2.
I calculated 2Ca to .171mol and O2 to .286mol, I am lost from here...
2 answers
Sweet. Thank you. I was real close. I was just calc ing the moles wrong at the end.
Thanks again
Thanks again