Question

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

Unbalanced equation: Na + S → Na2S

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

(I've trying looking it up and in my notes but I'm v confused)

Answer 1

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

Answer 2

I"ve been working on it and I think I got it right.

Balanced: 2Na + S → Na2S

moles Na available= 45.3 g/23 g/mole=1.97 mole

moles sulfur available = 105g/32.1 g/mole=3.27 mole

Moles Na2S = 1/2 (1.97)= 0.985 moles

Mass Na2S that can be produced=
0.985 moles (2*23 + 32.1g/mole)= 76.9 g

Answer 3

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 4

Ohh!! I see now. Thank you so much

Answer 5

in one mol of sodium there are 23 grams, and in one mol of sulfur there are 32.07 grams
but look what given to you, 2 moles of sodium which is 23X2 =46 grams which is ok but look how much sulfur is given to you 105g which is wayyyy more than you need cuz you only have 1 mol of sulfur and that's why sulfur si the excess reagent

Answer 6

sorry for "45.3g X 23g/mol = 1.96 mol of Na"
it should be 45.3g X 1mol/23g, where grams are cancels out and you are left with 1.96 mol of NA