What mass of Ag2s is produced from a mixture of 2.0g Ag and 2.0g S8?

Answer 1

The way to solve these problems is to determine which reactant will produce the fewest moles of product. In order to do this, we first need to convert the mass of each reactant into moles of reactant using the molecular mass of silver, Ag (107.9 g/mol), and molecular sulfur, S8 (256.5) g/mol:

2.0 g Ag * 1mol/107.9 g/mol = 0.0185 mol Ag

2.0 g S8 * 1mol/256.5 g/mol = 0.00779 mol S8

Now using the balanced chemical equation (our recipe) from above:
16 Ag + S8 -->8 Ag2S
We can see:
16 moles of Ag turns into 8 moles of Ag2S
1 mole of S8 turns into 8 moles of Ag2S

What we need to do know, is use these two statements above to see how many moles of product each reactant turns into:

0.0185 mol Ag * 8 mol Ag2S/16 mol Ag = 0.00925 mol Ag2S
0.00779 mol S8 * 8mol Ag2S/1 mol S8 = 0.0623 mol Ag2S

Since 0.0185 mole of Ag produce less Ag2S (the product), silver (Ag) is the limiting reactant. Now you could convert this to grams of product by multiplying the lesser amount produced, 0.00925 moles of Ag2S, by the molecular mass of Ag2S.

To determine the mass of reactant left unreacted, we need to convert moles of the limiting reactant into moles of the other reactant to find out how much of the other reactant is lost--Then subtract from the original amount and convert to grams with the molecular mass.

0.0185 mol Ag * 1 mol S8/16 mol Ag = 0.00116 mol S8
Originally we had 0.00779 mol of S8, from above, so:

0.00779 mol S8 - 0.00116 mol S8 = 0.00663 mol S8
Now convert to grams left:
0.00663 mol S8 * 256.5 g/mol = 1.70 g S8 left unreacted!