2.16 grams of silver react with 3.20 grans of sulfur.Caculate the mass of silver sulfide produced
(mass of ag=108.8g/mole)
You made a typo; the atomic mass of Ag is 107.8(rounded). This is a limiting reagent (LR) problem and the way you know that is because amounts are given for BOTH reactants.
2Ag + S ==> Ag2S
mols Ag = 2.16/atomic mass Ag = ?
mols S = grams/atomic mass S = ?
Using the coefficients in the balanced equation, convert mols Ag to mols Ag2S.
Do the same and convert mols S to mols Ag2S.
It is likely that the two values for mols Ag2S will not be the same which means one is wrong. The correct value in LR problems is ALWAYS the smaller value and the reagent responsible for that lower values is called the LR.
Now take the smaller number and convert to grams Ag2S.
mols Ag2S x molar mass Ag2S = grams Ag2S.
To calculate the mass of silver sulfide produced, you need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed and limits the amount of product that can be formed. To find the limiting reactant, you compare the number of moles of each reactant using their molar masses.
1. Calculate the number of moles of each reactant:
Number of moles of silver (Ag) = mass of Ag / molar mass of Ag
Number of moles of sulfur (S) = mass of S / molar mass of S
Given:
- Mass of Ag = 2.16 grams
- Molar mass of Ag = 108.8 g/mol
- Mass of S = 3.20 grams
- Molar mass of S = 32.06 g/mol
Number of moles of Ag = 2.16 g / 108.8 g/mol = 0.0199 mol
Number of moles of S = 3.20 g / 32.06 g/mol = 0.0999 mol
2. Determine the limiting reactant:
To find the limiting reactant, compare the moles of both reactants. The reactant with fewer moles is the limiting reactant.
In this case, silver (Ag) has fewer moles (0.0199 mol) compared to sulfur (S) (0.0999 mol). So, silver (Ag) is the limiting reactant.
3. Calculate the moles of product (silver sulfide, Ag2S):
The balanced chemical equation for the reaction between silver and sulfur is:
2Ag + S → Ag2S
From the balanced equation, we can see that 2 moles of Ag react with 1 mole of S to form 1 mole of Ag2S.
Since silver (Ag) is the limiting reactant, the moles of Ag2S formed will be equal to half the moles of Ag.
Moles of Ag2S = (0.0199 mol Ag) / 2 = 0.00995 mol
4. Calculate the mass of Ag2S produced:
Mass of Ag2S = moles of Ag2S x molar mass of Ag2S
Given:
- Molar mass of Ag2S = 247.8 g/mol
Mass of Ag2S = 0.00995 mol x 247.8 g/mol ≈ 2.46 grams
Therefore, the mass of silver sulfide (Ag2S) produced is approximately 2.46 grams.