How many grams of Ag2S can be generated from a mixture of 2.48 g Na2S and 3.41 g of AgNO3 given the following unbalanced reaction:
Na2S(aq) + AgNO3(aq) → Ag2S(aq) + NaNO3(aq)
moles = g/molar mass
moles AgNO3 = 3.41/approx 170 = approx 0.02
moles Na2S = 2.48/about 78 = about 0.032
.........2AgNO3 + Na2S ==>Ag2S + 2NaNO3
initial...0.02....0.032......0.....0
Now we take them one at a time.
First, 0.02 mol AgNO3 will produce how much Ag2S if we had all of the Na2S we needed. That will be 0.02 x (1 mol Ag2S/2 mol AgNO2) = 0.02 x 1/2 = about 0.01 mol Ag2S
How much will 0.032 mol Na2S produce if we have all of the AgNO3 we need. That is
0.032 x (1 mol Ag2S/1 mol Na2S) = about 0.032 x 1/1 = about 0.032 moles.
Obviously both answers can't be correct and in limiting reagent problems the smaller number is ALWAYS the correct one. So the maximum amount of Ag2S we can produce from this mixture is 0.01 mol. Convert that to grams by g = moles x molar mass if you want grams. The limiting reagent in this case is AgNO3 and there will be some Na2S remaining without reacting. All of the AgNO3 will be used and there will be none of that remaining.
thanks!
To find out how many grams of Ag2S can be generated from the given mixture, we need to follow these steps:
Step 1: Write the balanced chemical equation:
Na2S(aq) + AgNO3(aq) → Ag2S(aq) + 2NaNO3(aq)
Step 2: Calculate the molar mass of Na2S:
Na: 22.99 g/mol
S: 32.07 g/mol
Molar Mass of Na2S = (22.99 x 2) + 32.07 = 46 + 32.07 = 78.07 g/mol
Step 3: Calculate the number of moles of Na2S:
Moles = Mass / Molar Mass
Moles of Na2S = 2.48 g / 78.07 g/mol = 0.0318 mol
Step 4: Use the stoichiometric ratio to determine the number of moles of Ag2S that can be generated:
From the balanced equation: 1 mol of Na2S produces 1 mol of Ag2S
So, 0.0318 mol of Na2S will produce 0.0318 mol of Ag2S.
Step 5: Calculate the molar mass of Ag2S:
Ag: 107.87 g/mol
S: 32.07 g/mol
Molar Mass of Ag2S = (107.87 x 2) + 32.07 = 215.74 + 32.07 = 247.81 g/mol
Step 6: Calculate the mass of Ag2S:
Mass = Moles x Molar Mass
Mass of Ag2S = 0.0318 mol x 247.81 g/mol = 7.88 g
Therefore, 7.88 grams of Ag2S can be generated from the given mixture of 2.48 g Na2S and 3.41 g AgNO3.
To determine the number of grams of Ag2S that can be generated, we need to follow a few steps:
Step 1: Write the balanced equation for the reaction:
Na2S(aq) + 2AgNO3(aq) → Ag2S(aq) + 2NaNO3(aq)
Step 2: Calculate the molar masses of Na2S and AgNO3:
- Molar mass of Na2S = 2(22.99 g/mol) + 32.07 g/mol = 46.0 g/mol
- Molar mass of AgNO3 = 107.87 g/mol + 3(16.00 g/mol) = 169.87 g/mol
Step 3: Use the molar masses to convert the given masses of Na2S and AgNO3 to moles:
- Moles of Na2S = 2.48 g / 46.0 g/mol = 0.054 mol
- Moles of AgNO3 = 3.41 g / 169.87 g/mol = 0.0201 mol
Step 4: Identify the limiting reactant. The limiting reactant is the one that is completely consumed and determines the amount of product that can be formed. In this case, the reactant that produces the least amount of product is the limiting reactant.
- The balanced equation shows a 1:2 mole ratio between Na2S and Ag2S. Therefore, 1 mole of Na2S will produce 2 moles of Ag2S.
- From the mole calculations, we have 0.054 mol of Na2S, which would produce 2 * 0.054 mol = 0.108 mol of Ag2S.
- Similarly, 0.0201 mol of AgNO3 would produce 0.0201 mol of Ag2S.
Since the amount of Ag2S that can be formed is limited by the amount of Na2S, 0.108 mol of Ag2S can be generated.
Step 5: Convert the moles of Ag2S to grams:
- Grams of Ag2S = moles of Ag2S * molar mass of Ag2S
- Grams of Ag2S = 0.108 mol * 247.8 g/mol (molar mass of Ag2S)
- Grams of Ag2S = 26.78 g
Therefore, from the given mixture of 2.48 g Na2S and 3.41 g AgNO3, the maximum amount of Ag2S that can be generated is 26.78 grams.