The reaction will produce 12.1 g Ag₂S.
<em>Balanced equation</em> = 2Ag + S ⟶ Ag₂S
<em>Mass of Ag₂S</em> = 10.5 g Ag × (1 mol Ag/107.87 g Ag) × (1 mol Ag₂S/2 mol Ag)
× (247.80 g Ag₂S/1 mol Ag₂S) = 12.1 g Ag₂S
Answer:
B I think I am pretty sure
Answer:
Explanation:
To solve this problem, we need to obtain the number of moles of the solute we desired to prepare;
Number of moles = molarity x volume
Parameters given;
volume of solution = 500mL = 0.5L
molarity of solution = 0.5M
Number of moles = 0.5 x 0.5 = 0.25moles
Now to know the volume stock to take;
Volume of stock =
molarity of stock = 4M
volume =
= 0.0625L or 62.5mL