For the reaction, calculate the moles of indicated product produced when 5.00 moles of each reactant is used.
4Li(s)+O2(g)+ < 2Li2O(s) (Li2O)
3 answers
You need to explain the problem. I'm sure the parentheses mean something but I don't know what. And do you realize you have the arrow pointing backwards? Does that mean anything.
okay. the Li2O in parenthesis means that the Li2O is the reactant used. So, I am looking for how to calculate the moles of indicated product produced in the equation when 5.00 moles of Li20 is used. The < means the sign for the reaction arrow I thought? I am not sure how to type it out. Thanks!
OK. You didn't say it this way but I THINK, from the wording in the problem, that you have 5.00 mols Li and 5.00 mols O2 (those are the reactants) and you want to know how much Li2O (the indicated product) is formed. To convert mols of one thing to mols of another, you use the coefficients in the balanced equation. For 5.00 mols Li, that is
mols Li2O = 5.00 mols Li x (2 mols Li2O/4 mol Li) = 5.00 x (2/4) = 2.50 mols Li2O.
For 5.00 mols O2 converted to mols Li2O, it is
mols Li2O = 5.00 mols O2 x (2 mols Li2O/1 mol O2) = 5.00 x (2/1) = 10.0 mols Li2O.
If I have misinterpreted the question, I think with this information you can convert anything in the equation to any other item. For example, if I want to convert 5.00 mols Li2O to mols O2 it is
5.00 mols Li2O x (2 mols O2/1 mol Li2O) = 2.50 mols O2. No matter what numbers you pick for O2, mols Li2O will always be twice O2. mols Li2O will always be 1/2 mols Li. And Li will always be twice Li2O and O2 will be 1/2 Li2O.
mols Li2O = 5.00 mols Li x (2 mols Li2O/4 mol Li) = 5.00 x (2/4) = 2.50 mols Li2O.
For 5.00 mols O2 converted to mols Li2O, it is
mols Li2O = 5.00 mols O2 x (2 mols Li2O/1 mol O2) = 5.00 x (2/1) = 10.0 mols Li2O.
If I have misinterpreted the question, I think with this information you can convert anything in the equation to any other item. For example, if I want to convert 5.00 mols Li2O to mols O2 it is
5.00 mols Li2O x (2 mols O2/1 mol Li2O) = 2.50 mols O2. No matter what numbers you pick for O2, mols Li2O will always be twice O2. mols Li2O will always be 1/2 mols Li. And Li will always be twice Li2O and O2 will be 1/2 Li2O.