Question

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

Answer 1

Answer:

$2.03~g~Ag_2S$

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 $H_2S$:

$0.280~g~H_2S\frac{34~g~H_2S}{1~mol~H_2S}=0.0082~mol~H_2S$

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:

$0.0082~mol~H_2S\frac{2~mol_Ag_2S}{2~mol~H_2S}=0.0082~mol~H_2S$

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

$0.0082~mol~H_2S\frac{247.8~g~Ag_2S}{1~mol~H_2S}=2.03~g~Ag_2S$

I hope it helps!