Sodium sulfide that can be produced from 45.3 g of sodium reacting with 105 g of sulfur is 76.8 g. The balanced chemical equation for the reaction between sodium and sulfur is 2Na(s) + S(s) → Na2S(s). Using this equation and mole calculations based on the masses provided, sodium is the limiting reagent and will produce 0.985 moles of sodium sulfide, which corresponds to a mass of 76.8 g.

1. 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.
Solution
Balanced equation,
2 Na (s) + S (s) ------------> Na2S ( s )
Moles of Na = mass / molar mass = 45.3 / 23 = 1.97 mol
Moles of S = 105 / 32 = 3.28 mol
From the balanced equation,
2 mol of Na needs 1 mol of S
Then, 1.97 mol of Na needs 0.985 mol of S,
So, Na is limiting reagent.
From the balanced equation,
2 mol of Na forms 1 mol of Na2S
then,
1.97 mol of Na forms 1.97 * 1 / 2 = 0.985 mol of Na2S
Mass of Na2S formed = moles * molar mass = 0.985 * 78 = 76.8 g.