FeCl3(aq) + K2S(aq) Fe2S3(aq) + KCl(aq)
Consider the unbalanced equation above.
What is the maximum mass of Fe2S3 that can be produced when 54.5 mL of 0.480 M FeCl3 and 98.5 mL of 0.420 M K2S react?
This is a limiting reagent (LR) problem. I know that because amounts are given for BOTH reactants. First, balance equation.
2FeCl3(aq) + 3K2S(aq) ==> Fe2S3(aq) + 6KCl(aq)
mols FeCl3 = M x L = ?
mols K2S = M x L = ?
Using the coefficient in the balanced equation, convert mols FeCl3 to mols Fe2S3.
Do the same for mols K2S to mols Fe2S3.
It is likely that these two value will not agree which means one of them is wrong; the correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is the LR.
Now convert mols Fe2S3 to grams. g = mols x molar mass.
To find the maximum mass of Fe2S3 that can be produced, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed and determines the amount of product formed.
Let's start by calculating the number of moles of FeCl3 and K2S:
Number of moles of FeCl3 = volume (L) x concentration (mol/L)
= 0.0545 L x 0.480 mol/L
= 0.02616 mol
Number of moles of K2S = volume (L) x concentration (mol/L)
= 0.0985 L x 0.420 mol/L
= 0.04137 mol
Next, we need to find the stoichiometric ratio of FeCl3 to Fe2S3 from the balanced equation:
FeCl3(aq) + K2S(aq) Fe2S3(aq) + KCl(aq)
1 mole of FeCl3 produces 1 mole of Fe2S3
Now, let's calculate the moles of Fe2S3 that can be formed from FeCl3:
Moles of Fe2S3 = Moles of FeCl3
= 0.02616 mol
Similarly, let's calculate the moles of Fe2S3 that can be formed from K2S:
Moles of Fe2S3 = Moles of K2S x (1 mole of Fe2S3 / 1 mole of K2S)
= 0.04137 mol x (1 mole of Fe2S3 / 1 mole of K2S)
= 0.04137 mol
Since both reactants produce the same number of moles of Fe2S3, FeCl3 is the limiting reactant.
Now, we can calculate the mass of Fe2S3 using its molar mass:
Molar mass of Fe2S3 = 55.85 g/mol (molar mass of Fe) + 32.07 g/mol (molar mass of S)
= 87.92 g/mol
Mass of Fe2S3 = Moles of Fe2S3 x Molar mass of Fe2S3
= 0.02616 mol x 87.92 g/mol
= 2.29 g
Therefore, the maximum mass of Fe2S3 that can be produced is 2.29 grams.
To find the maximum mass of Fe2S3 that can be produced, we first need to balance the chemical equation that describes the reaction. The balanced equation for the reaction between FeCl3 and K2S is:
2FeCl3(aq) + 3K2S(aq) -> Fe2S3(aq) + 6KCl(aq)
From the balanced equation, we can see that 2 moles of FeCl3 react with 3 moles of K2S to produce 1 mole of Fe2S3.
Now, let's calculate the number of moles of FeCl3 and K2S given the volumes and molarities provided.
Volume of FeCl3 = 54.5 mL = 0.0545 L
Molarity of FeCl3 = 0.480 M
Number of moles of FeCl3 = volume x molarity = 0.0545 L x 0.480 M = 0.0262 moles
Volume of K2S = 98.5 mL = 0.0985 L
Molarity of K2S = 0.420 M
Number of moles of K2S = volume x molarity = 0.0985 L x 0.420 M = 0.0413 moles
According to the balanced equation, the ratio of FeCl3 to Fe2S3 is 2:1. Therefore, since we have 0.0262 moles of FeCl3, we can produce half that amount of Fe2S3, which is 0.0262 / 2 = 0.0131 moles.
Now, let's find the molar mass of Fe2S3:
Molar mass of Fe = 55.85 g/mol
Molar mass of S = 32.06 g/mol
Total molar mass of Fe2S3 = 2 x (55.85 g/mol) + 3 x (32.06 g/mol) = 207.52 g/mol
Finally, we can calculate the maximum mass of Fe2S3 produced:
Mass of Fe2S3 = number of moles x molar mass
= 0.0131 moles x 207.52 g/mol
= 2.720 g
Therefore, the maximum mass of Fe2S3 that can be produced is 2.720 grams.