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.
THIS IS WHAT I PUT.. is this correct? someone please let me know!
and thank you!
88.1 g S x (247.8gmol/ 32.065) = 680.84
Ag 2 = 2*107.87 = 215.74 grams / mol of Ag2S
S = 32.06 grams / mol of Ag2S
so
Ag2S = 247.8 grams / mol ofAg2S
so yes
Thank you so much!
You are welcome.
To solve this problem correctly, we need to use the molar mass of sulfur (S) and the molar mass of silver sulfide (Ag2S) to convert between grams and moles.
First, let's calculate the number of moles of sulfur (S) in 88.1 grams:
88.1 g S x (1 mol S / 32.065 g S) = 2.747 mol S
Next, we use the balanced chemical equation for the reaction:
Ag2S + O2 → Ag + SO2
According to the equation, one mole of Ag2S produces two moles of Ag. Therefore, the number of moles of Ag2S can be calculated by dividing the moles of sulfur (S) by 2:
2.747 mol S / 2 = 1.374 mol Ag2S
Finally, we need to calculate the grams of Ag2S. To do this, we multiply the moles of Ag2S by its molar mass:
1.374 mol Ag2S x (247.8 g Ag2S / 1 mol Ag2S) = 340.0 g Ag2S
Therefore, the correct answer is 340.0 g Ag2S.