If 216 grams of silver are mixed with 40 grams of sulfur, how much silver sulfide, in grams, will be produced if there are 8 grams of sulfur left at the end of the reaction?

Question

In the reaction to create silver sulfide, it reacts with sulfur. If 216 grams of silver are mixed with 40 grams of sulfur, how much silver sulfide, in grams, will be produced if there are 8 grams of sulfur left at the end of the reaction?

Answer 1

For this we need to write a balanced equation. To begin, let's write down what we know:

Ag(s) + S(s) -> Ag₂S(s)

This equation is unbalanced. Therefore, the balanced equation needs to be

2Ag(s) + S(s) -> Ag₂S(s)

Now, we use dimensional analysis to solve the problem. Because there are 8 grams of Sulfur left over, this implies that 32 grams of sulfur are in the product.

32 g Sulfur x (1 mole S/32.065 g S) X (1 mole Ag₂S/1 mol S) x (247.8g / 1 mol Ag₂S) = 247.29 g.

Note: the answer is not in significant figures. If you need help with that, let me know.