# Ag(s) + H2S(g) + O2(g) Ag2S(s) + H2O(l) Consider the unbalanced equation above. What is the maximum mass of?

Ag2S that can be formed when 0.650 g of Ag, 0.165 g H2S and 0.255 g O2 react? If you could use significant figures, that would be awesome! Thank you :))

### 2 Answers

- kanshouLv 47 years agoFavorite Answer
4 Ag(s) + 2 H2S(g) + O2(g) ==> 2 Ag2S(s) + 2 H2O(l)

0.650 g Ag / 107.87 g/mole Ag = 0.0060257718 moles of Ag x 2/4 molar ratio Ag2S/Ag =

0.0030128859 moles of Ag2S

0.165 g H2S / 34.07 g/mole H2S = 0.0048429704 moles of H2S x 2/2 molar ratio Ag2S/H2S =

0.0048429704 moles of Ag2S

0.255 g O2 / 32.00 g/mole O2 = 0.00796875 moles of O2 x 2/1 molar ratio Ag2S/O2 =

0.0159375 moles of O2

Ag is the limiting reagent

0.0030128859 moles of Ag2S x 247.79 g/mole Ag2S = 0.7465629972 g Ag2S

0.747 g to 3 S.F.

We were taught to never round until the end or you could compromise accuracy. 0.00301 x 247.79 = 0.7458479 or 0.746 to 3 S.F. so there is a slight difference.

- wiedemannLv 43 years ago
i dont have a calculator with me so...right that is a precise social gathering, basically stick with it, you're going to get g it, its the picture of doing it for you. Balanced equation: 4Ag(s) + 2H?S(g) + O?(g) ---> 2Ag?S(s) + 2H?O(l) locate the proscribing reactant. For Ag: reactant = mole / coefficient reactant = (0.930 g / 107.87 g/mol) / 4 reactant = 0.002155 For H?S: reactant = mole / coefficient reactant = (0.one hundred fifteen g / 34.08 g/mol) / 2 reactant = 0.001687 For O?: reactant = mole / coefficient reactant = (0.four hundred g / 32.00 g/mol) / a million reactant = 0.0125 considering that 0.001687 is below 0.0125 and nil.002155, then H?S is the proscribing reactant. we can now locate the optimal mass of silver sulfide which will form. in this formula, the product (Ag?S) takes x, and the proscribing reactant takes y. (? = coefficient contained in the balanced equation) mass of x = (mole of y) * (? mol x / ? mol y) * (molar mass of x) mass of Ag?S = (0.one hundred fifteen g / 34.08 g/mol) * (2 mol / 2 mol) * (247.80 one g/mol) mass of Ag?S = 0.836 g (answer)