In the reaction to create silver sulfide, silver reacts with sulfur, as described by this equation: 2Ag + S -- Ag2S. If
215.8 grams of silver is mixed with 32.1 grams of sulfur, how much silver sulfide, in grams, will be produced? Use
equations to help you solve for the answer.

247.9 g

Explanation:

First, convert mass to moles.

215.8 g Ag (1 mol Ag / 107.9 g Ag) = 2.00 mol Ag

32.1 g S (1 mol S / 32.1 g S) = 1.00 mol S

Find the limiting reactant (if there is one):

2.00 mol Ag (1 mol S / 2 mol Ag) = 1.00 mol S

There is no limiting reactant (both reactants completely react).  So use either reactant to find moles of product.

2.00 mol Ag (1 mol Ag₂S / 2 mol Ag) = 1.00 mol Ag₂S

Convert moles to mass:

1.00 mol Ag₂S (247.9 g Ag₂S / mol Ag₂S) = 247.9 g Ag₂S

c. electrical charges of ionic bonds.

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