Question

How many grams of silver sulfide are formed when 0.280 g of hydrogen sulfide reacts with excess silver and oxygen?

Answer 1

`2.03~g~Ag_2S`

Step-by-step explanation:

For this question we have to start with the reaction:

`Ag~+~H_2S~+~O_2~->~Ag_2S~+~H_2O`

Now, we can balance the reaction, so:

`4Ag~+~2H_2S~+~O_2~->~2Ag_2S~+~2H_2O`

With this in mind, we have to start with the amount of
`H_2S`. The first step is to convert from grams to moles. For this, we need to find the molar mass of
`H_2S`. If we check the periodic table we will find the atomic masses for Ag and H; H: 1 g/mol and A: 32 g/mol, so:

(1*2)+ (32*1) = 34 g/mol.

Now we can calculate the moles of
`H_2S`:

`0.280~g~H_2S(34~g~H_2S)/(1~mol~H_2S)=0.0082~mol~H_2S`

With the moles of
`H_2S` we can calculate the moles of
`Ag_2S` if we check the molar ratio in the balanced equation,
`2~mol~H_2S=2~mol_Ag_2S`, so:

`0.0082~mol~H_2S(2~mol_Ag_2S)/(2~mol~H_2S)=0.0082~mol~H_2S`

With the molar mass of
`Ag_2S` we can convert from moles to grams (Ag: 107.86 g/mol, S: 32 g/mol), so:

(107.86*1)+(32*2)=247.80 g/mol

`0.0082~mol~H_2S(247.8~g~Ag_2S)/(1~mol~H_2S)=2.03~g~Ag_2S`

I hope it helps!