posted by Annie .
Zinc and sulfur react to form zinc sulfide according to the equation Zn+S=>ZnS
a) How many grams of ZnS will be formed?
b) How many grams of the excess reactant will remain after the reaction is over?
How much is formed depends upon how much you start with. My crystal ball isn't working that well tonight.
Oh, sorry. 25 g of Zinc and 30 g of sulfur. I already found that Zinc is the limiting reagent, but I'm stuck on the rest.
If you know Zn is the limiting reagent then you must have moles Zn calculated as about 0.38.
Now you want to convert moles Zn to mols ZnS. To do that you use the coefficients in the balanced equation like this.
0.38 mol Zn x (1 mol ZnS/1 mol Zn) = 0.38 mol Zn x 1/1 so 0.38 mols Zn = 0.38 mol ZnS. You can convert that to grams by mols x molar mass = grams. (I just estimated mols Zn; I'm sure you have a better figure for that.)
To find S remaining S, do this. You have moles Zn. Convert to moles S (same process as converting mols Zn to mols ZnS. That will tell you how many moles S were used by the 25 g Zn. Then mols S x atomic mass S = grams S and that subtracted from 32 will tell you how many grams S did not react.