4 Ag + 2 H2S + O2 = 2 Ag2S + 2 H2O

a) What mass if silver sulfide is formed from the reaction of 0.120g of silver?
b) What mass of hydrogen sulfide is needed in the same process?

So I understand a, and I found the answer to be 0.138g. For b the text book says the answer is 0.0190g. I don't understand how they got to that answer because they never tell us the mass of the hydrogen sulfide

I'd really appreciate your help, thank you (:

Convert given data to moles, relate to the reaction ratios, then convert back to grams.

a) for 0.12g Ag = 0.0011mole Ag => produces 1/2(0.0011mole) Ag2S => (0.00056mole)(248g/mole) = 0.138g Ag2S

b) assume an excess of H2S is present, then the rxn will require at least 1/2(moles Ag used). From (a) moles Ag = 0.0011mole then => moles H2S required is 1/2(0.0011)mole = 0.00056mole H2S = (0.00056mole)(34g/mole)H2S = 0.019g H2S

To solve part (b) of the question, you are correct that we are not given the mass of hydrogen sulfide (H2S) directly. However, we can determine it using stoichiometry and the information provided in the balanced chemical equation.

From the balanced equation: 4 Ag + 2 H2S + O2 = 2 Ag2S + 2 H2O

We can see that 4 moles of silver react with 2 moles of hydrogen sulfide to produce 2 moles of silver sulfide. The mole ratio is 4:2, which simplifies to 2:1 or 2 moles of Ag to 1 mole of H2S.

Since we have the mass of silver as 0.120g, we can convert it to moles using the molar mass of silver (taken from the periodic table). The molar mass of Ag is 107.87 g/mol.

0.120g Ag * (1 mol Ag / 107.87g Ag) = 0.001114 mol Ag

Since the mole ratio of Ag to H2S is 2:1, we can determine the amount of H2S required by doubling the moles of silver:

0.001114 mol Ag * (1 mol H2S / 2 mol Ag) = 0.000557 mol H2S

Now, we can convert the moles of H2S to grams using its molar mass (again, taken from the periodic table). The molar mass of H2S is 34.08 g/mol.

0.000557 mol H2S * (34.08 g H2S / 1 mol H2S) = 0.019g H2S

Therefore, the mass of hydrogen sulfide required for the reaction is 0.0190g (rounded to four decimal places). Your textbook's answer of 0.0190g is correct.

To solve part b of the question, we need to find the mass of hydrogen sulfide (H2S) needed in the reaction. Although the mass of H2S is not directly given in the question, we can use the stoichiometric coefficients to determine the ratio between the reactants.

In the balanced chemical equation:
4 Ag + 2 H2S + O2 = 2 Ag2S + 2 H2O

We can see that the ratio between Ag and H2S is 4:2 or 2:1. This means that for every 2 moles of Ag, we need 1 mole of H2S.

Since we know the mass of silver (0.120g) in part a, we can first find the number of moles of silver using its molar mass.

Ag molar mass: 107.87 g/mol
moles of Ag = mass of Ag / molar mass of Ag
moles of Ag = 0.120g / 107.87 g/mol
moles of Ag = 0.001113 mol (approximately)

Now, using the mole ratio from the balanced equation, we can find the moles of H2S required.

moles of H2S = moles of Ag / 2
moles of H2S = 0.001113 mol / 2
moles of H2S = 0.0005565 mol (approximately)

Finally, we can calculate the mass of H2S using its molar mass.

H2S molar mass: 34.08 g/mol
mass of H2S = moles of H2S * molar mass of H2S
mass of H2S = 0.0005565 mol * 34.08 g/mol
mass of H2S = 0.018982 g (approximately)

So, the mass of hydrogen sulfide (H2S) needed in the reaction is approximately 0.0190 g, as stated in the textbook answer.