How many grams of Ag2S react to form 174 g of AgCl? see below:
Ag2S + 2 HCl ==> 2 AgCl + H2S
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175g AgCl x (1 mol/247.9g) x (1 Ag2S/2AgCl) x (143.4g/1 mol) = 50.615g Ag2S
I'm assuming you've learned about basic stoichiometry and how to use mole bridges? Otherwise this type of question wouldn't be asked.
To determine the number of grams of Ag2S that react to form 174 g of AgCl, we need to use the balanced equation and calculate the molar masses of Ag2S and AgCl.
First, let's calculate the molar mass of Ag2S:
Ag2S:
Ag: atomic mass = 107.87 g/mol
S: atomic mass = 32.07 g/mol
Molar mass of Ag2S = (2 * Ag) + S
= (2 * 107.87 g/mol) + 32.07 g/mol
= 355.81 g/mol
Next, let's calculate the molar mass of AgCl:
AgCl:
Ag: atomic mass = 107.87 g/mol
Cl: atomic mass = 35.45 g/mol
Molar mass of AgCl = Ag + Cl
= 107.87 g/mol + 35.45 g/mol
= 143.32 g/mol
Now, we can use the molar mass and the balanced equation to solve for the grams of Ag2S.
The balanced equation shows that 1 mole of Ag2S reacts with 2 moles of HCl to produce 2 moles of AgCl. Therefore, the molar ratio between Ag2S and AgCl is 1:2.
To calculate the grams of Ag2S needed, we use the following equation:
grams of Ag2S = (grams of AgCl) * (molar mass of Ag2S/molar mass of AgCl)
Substituting the given value:
grams of Ag2S = 174 g * (355.81 g/mol / 143.32 g/mol)
Calculating the expression inside the parentheses:
355.81 g/mol / 143.32 g/mol ≈ 2.48
Multiplying the molar mass ratio by the given grams of AgCl:
grams of Ag2S = 174 g * 2.48
grams of Ag2S ≈ 431.52 g
Therefore, approximately 431.52 grams of Ag2S react to form 174 grams of AgCl.