Argentite (Ag2S) is one of the common ores of silver. If 88.1 grams of sulfur was removed from ore, how many grams of Ag2S were mined to get pure silver?
_________ g Ag2S. Do NOT enter the unit and report your final answer with 4 SFs.
mm = molar mass
88.1 g S x (mm Ag2S/atomic mass S) = ? g Ag2S
@DrBob222
then it would be something like this.. 88.1g S x (247.9/32.065) =
To find the grams of Ag2S that were mined, we need to calculate the molar mass of Ag2S and then use stoichiometry to determine the amount of Ag2S.
Step 1: Calculate the molar mass of Ag2S.
The molar mass of Ag2S is the sum of the atomic masses of all the atoms in Ag2S.
Ag (silver) has an atomic mass of 107.87 g/mol
S (sulfur) has an atomic mass of 32.07 g/mol
Molar mass of Ag2S = (2 * Ag) + S
= (2 * 107.87 g/mol) + 32.07 g/mol
= 215.74 g/mol + 32.07 g/mol
= 247.81 g/mol
Step 2: Use stoichiometry to calculate the amount of Ag2S mined.
We know that 88.1 grams of sulfur was removed from the ore. The molar mass of Ag2S is 247.81 g/mol. We can use these values to calculate the amount of Ag2S.
Amount of Ag2S = (mass of sulfur removed) / (molar mass of Ag2S)
= 88.1 g / 247.81 g/mol
= 0.35507 mol
Step 3: Convert the amount of Ag2S to grams.
To convert the amount of Ag2S from moles to grams, we multiply the amount by the molar mass of Ag2S.
Grams of Ag2S = (amount of Ag2S) * (molar mass of Ag2S)
= 0.35507 mol * 247.81 g/mol
= 87.97 g
Therefore, the number of grams of Ag2S mined to get pure silver is 87.97 g.