This is a limiting reagent problem (LR) , and a difficult one at that, and the only way I know that you can be SURE you have it correct is to do it the long way. Basically that is to determine how many grams Ag2S formed from each of the reactants and to pick the smallest one. I hope oobleck shows you the short way.
4Ag(s) + 2H2S(g) + O2(g) → 2Ag2S(s)+ 2H2O(g)
.1.90 g.....0.280 g.....0.160 g
mols Ag = 1.90 g/107.9 = 0.0176
mols H2S = 0.280/34 = 0.00823
mols O2 = 0.160/32 = 0.005
Using the coefficients convert each reactant, by itself, to moles of product of Ag2S.
For Ag: 0.0176 moles Ag x (2 mols Ag2S/4 mols Ag) = 0.0088
For H2S: 0.00823 mols H2S x (2 moles Ag2S/2 mols H2S) = 0.00823
For O2: 0.005 mols O2 x (2 moles Ag2S/1 mol O2) = 0.01
The smallest number of moles Ag2S formed is from the H2S; therefore, 0.00823 mols Ag2S will be formed. Convert that to grams. g = mols x molar mass. Post your work if you get stuck.
Silver reacts with hydrogen sulphide gas, and oxygen according to the reaction:
4Ag(s) + 2H2S(g) + O2(g) → 2Ag2S(s)+ 2H2O(g)
How many grams of silver sulphide are formed when 1.90 g of silver reacts with 0.280 g of
hydrogen sulphide and 0.160 g of oxygen?
1 answer