moles = g/molar mass
moles AgNO3 = 3.41/approx 170 = approx 0.02
moles Na2S = 2.48/about 78 = about 0.032
.........2AgNO3 + Na2S ==>Ag2S + 2NaNO3
initial...0.02....0.032......0.....0
Now we take them one at a time.
First, 0.02 mol AgNO3 will produce how much Ag2S if we had all of the Na2S we needed. That will be 0.02 x (1 mol Ag2S/2 mol AgNO2) = 0.02 x 1/2 = about 0.01 mol Ag2S
How much will 0.032 mol Na2S produce if we have all of the AgNO3 we need. That is
0.032 x (1 mol Ag2S/1 mol Na2S) = about 0.032 x 1/1 = about 0.032 moles.
Obviously both answers can't be correct and in limiting reagent problems the smaller number is ALWAYS the correct one. So the maximum amount of Ag2S we can produce from this mixture is 0.01 mol. Convert that to grams by g = moles x molar mass if you want grams. The limiting reagent in this case is AgNO3 and there will be some Na2S remaining without reacting. All of the AgNO3 will be used and there will be none of that remaining.
How many grams of Ag2S can be generated from a mixture of 2.48 g Na2S and 3.41 g of AgNO3 given the following unbalanced reaction:
Na2S(aq) + AgNO3(aq) → Ag2S(aq) + NaNO3(aq)
2 answers
thanks!