Dawson H.

asked • 02/12/21

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

1 Expert Answer


Dawson H.

We have to type in our answers on some homework platform and it kept saying it was wrong. You only get three tries to type it in before it gives you a 0. But I'm glad you got what I got and showed me how to do C. I guess something is wrong with the platform.


J.R. S.

It may have to do with significant figures, so be sure to check that. You should have only 3 sig.figs. in your answer which is looks like you do. Sorry I can't offer any other suggestions at this time.


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