Question

How many grams of sodium sulfide can be produced when 45.3 g Na react with 105 g S?

Answer 1

This is what I have so far... This is the balanced equation. The limiting reactant is sodium because it will run out before sodium does. Therefore, sodium is the excess. First, I must how many moles are in the equation. From the balanced equation, I can find that the given amonts equal to a certain amount of moles. I can find that two moles of sodium is equal to 46 g 23g/mol 32.07g/mol 78.05g/mol.
45.3g X 23g/mol = 1.96 mol of Na

1.96 X 1 mol of Na2S/ 2 mol Na = 0.98mol of Na2S

0.98 mol of Na2S X 76.48g/mol where moles cancels out and you get 76.48 g of Na2S

Answer 2

https://vcsohio.brainhoney.com/MathML/Image.ashx?data=2Na%20%2B%20S%20-%3E%20Na_2S

Balanced equation

Answer 3

@fallenangelorchid

Answer 4

Write down the balance equation first....In this case i'll be
2Na(s) + S(s) = Na2S(s)

Answer 5

I have that. Check the link.

Answer 6

after that label what's already given to you
2Na(s) + S(s) = Na2S(s)
45.3g 105g m=?
23g/mol 32.07g/mol 78.05g/mol

in this case, the limiting reagent is Na and excess is Sulfur because sulfur is available in excess.

45.3g X 23g/mol = 1.96 mol of Na

1.96 X 1 mol of Na2S/ 2 mol Na = 0.98mol of Na2S

0.98 mol of Na2S X 76.48g/mol where moles cancels out and you get 76.48 g of Na2S

Answer 7

Bro, you just took that from another page, I need a different way, because I don't understand how they done it.

Answer 8

Kk then try this : https://answers.yahoo.com/question/index?qid=20101228170118AAU6NkK

Answer 9

Do you know how to do it or no?

Answer 10

@nevermind_justschool Not really bruh

Answer 11

It's alright, A for effort I guess LoL