Question

How many grams of silver nitrate are required to produce 0.25 mol of silver sulfide

Answer 1

To produce 0.25 mol of silver sulfide, you will need 42.47 grams of silver nitrate.

Step-by-step explanation:

To determine the number of grams of silver nitrate required to produce 0.25 mol of silver sulfide, we need to use the balanced equation of the reaction between silver nitrate and silver sulfide. According to the equation:

AgNO3 + Na2S → Ag2S + 2NaNO3

We see that 1 mole of AgNO3 reacts with 1 mole of Ag2S. Therefore, if we have 0.25 mol of Ag2S, we will need 0.25 mol of AgNO3. To convert that to grams, we multiply the number of moles by the molar mass of AgNO3:

0.25 mol AgNO3 * (169.88 g/mol) = 42.47 g AgNO3

Answer 2

62 grams.

In these problems start with the given and do dimensional analysis. I used the molar mass in the dimensional analysis.