Question

What is the limiting reactant when 8.4 moles of lithium react with 4.6 moles of oxygen gas?

Unbalanced equation: Li + O2 → Li2O

Show, or explain, all of your work along with the final answer.

Answer 1

@zendaya-replay-guy

Answer 2

yea

Answer 3

okay let me see

Answer 4

Balanced Eqn:

2Na+ S --> Na2S

45.3 g Na * 1moleNa/23gNa * 1mole NaS/2molNa * 78g/1molNaS = 76.81 grams NaS

Balanced eqn: 4Li + 02 ---> 2 Li2O
the limiting reactant is lithium becaus oxygen is in excess for every 4 moles of Li there needs to be 1 mole of Oxygen, since there is 8.4 mole Li there should be 4.2 moles , but oxygen is in excess thus Li it limiting the reaction so it can not proceed any further
Balanced Equation: 2Na + S --> Na2S

45.3g Na /22.99 g Na = 1.97 mol Na

1.97 mol Na/ 2 mol of Na --> 0.985 mol of Na2S

105 g S / 32.1 g S = 3.27 mol S --> 2.37 mol of Na2S

Since Na yields the least amount of Na2S, Na is the limiting reagent (LR). Therefore, 0.985 mols of Na2S are produced.

0.985 * 78.08 = 76.9g of Na2S is produced



2.) use steps in 1 to find the LR. To start, here is the balanced equation...

4Li + O2 --> 2Li2O

Answer 5

u still there