I keep getting lost on this question: In a combination reaction, 1.54 g of lithium is mixed with 6.56 g of oxygen.....

Question

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)+O_2(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

Answer 1

The question doesn't ask for the mass of oxygen consumed, but I believe all of your answers are correct, as they agree with what I have shown below. So, did someone tell you your answers were incorrect?

4Li + O₂ ==> 2Li₂O ... balanced equation

Next, find the limiting reactant. You can do this by finding moles of each reactant and dividing them by the coefficient in the balanced equation:

For Li: 1.54 g Li x 1 mol Li/6.94 g = 0.222 moles Li (÷4 -> 0.0555)

For O2: 6.56 g O2 x 1 mol O2/32 g = 0.205 moles O2 (÷1 -> 0.205)

(a) Since the value 0.0555 is less than 0.205, Li is the limiting reactant and will determine the moles of product that can be formed. So, O₂ is present in excess.

(b) 0.222 moles Li x 2 moles Li₂O / 4 moles Li = 0.111 moles Li₂O formed

(c) moles O₂ used up = 0.222 mol Li x 1 mol O₂ / 4 mol Li = 0.0555 mol O₂ used up

moles O₂ remaining = 0.205 mol - 0.0555 mol = 0.1495 mol O₂ remaining

mass O₂ remaining = 0.1495 mol O₂ x 32 g/mol = 4.78 g O₂ remaining

mass Li remaining = 0 g Li remaining

mass Li₂O formed = 0.111 moles Li₂O x 29.88 g Li₂O/mol = 3.32 g Li₂O formed