I keep getting lost on this question: In a combination reaction, 1.54 g of lithium is mixed with 6.56 g of oxygen.....
a) Which reactant is present in excess?
I got Lithium being the LR.
b) How many moles of the product are formed?
I got 3.32 g Li2O
c) After the reaction, how many grams of each reactant and product are present?
Blank g Li
Blank g O2
Blank g Li2O
I got 1.78 g O2 consumed.
I don't think any of my math is correct and I don't know how to answer c.
Here is my math so far:
BCE: 4Li(s)+O2(g) ------> 2Li2O(s)
1.54 g Li X 1 mol Li over 6.94 g Li = 0.222 mol Li
6.56 X 1 mol O2 over 32.00 g O2 = 0.205 mol O2
0.222 mol Li X 2 mol Li2O over 4 mol Li = 0.111 mol Li2O LR
0.205 mol O2 X 2 mol Li2O over 1 mol O2 = 0.41 mol Li2O
0.111 mol Li2O X 29.88 g Li2O over 1 mol Li2O = 3.32 g Li2O
0.222 mol Li X 1 mol O2 over 4 mol Li X 32.00 g O2 over 1 mol O2 = 1.78 g O2 consumed