If 120 mL of 2.8333 M aqueous HCl reacts stoichiometrically according to the balanced equation, how many milliliters of 4.43 M aqueous FeCl3 are produced?
Fe2S3(s) + 6HCl(aq) → 3H2S(g) + 2FeCl3(aq)
Molar Mass (g/mol)
HCl 36.461
FeCl3 162.21
Density (g/mL)
-
Molar Volume (L)
22.4 at STP
Gas Constant
(L.atm.mol-1.K-1)
0.0821
I did
(2.8333mol x 120mL x 2mol FeCl3)/ (1000ml x 6mol)
got .113332mols and converted to mL by
(4.43mol x 1000) / .113332mol
and got 39089ml
i don't know if i did it correctly, the number seem pretty big
I did
(2.8333mol x 120mL x 2mol FeCl3)/ (1000ml x 6mol)
got .113332mols and converted to mL by
You are right to here with 0.1133 moles FeCl3. Then
M = moles/L and rearrange to
L = moles/M = 0.1133/4.43 = 0.02558 L = 25.58 mL. However, there is a flaw in the problem. Since you started with 120 mL of HCl, the final volume of the solution will be 120 mL and the molarity will be 0.1133 moles/0.120 L = 0.944 M OR if you evaporated the aqueous solution until 25.58 mL remained, then the solution would be 4.43 M.
(4.43mol x 1000) / .113332mol
and got 39089ml
i don't know if i did it correctly, the number seem pretty big
To calculate the volume of 4.43 M aqueous FeCl3 produced, you correctly used the stoichiometry from the balanced equation. Here's the step-by-step process to verify your calculation:
1. Calculate the number of moles of HCl:
Using the given concentration of 2.8333 M and the volume of 120 mL:
Number of moles of HCl = concentration × volume / 1000
= (2.8333 mol/L) × (120 mL) / 1000
= 0.34 mol
2. Use the stoichiometry from the balanced equation to relate the moles of HCl to FeCl3:
From the balanced equation, 1 mole of Fe2S3 reacts with 6 moles of HCl to produce 2 moles of FeCl3.
So, the ratio of moles of HCl to moles of FeCl3 is 6:2 or 3:1.
Therefore, the number of moles of FeCl3 = 0.34 mol × (1 mol FeCl3 / 3 mol HCl)
= 0.1133 mol
3. Convert the moles of FeCl3 to milliliters:
Using the given concentration of 4.43 M:
Volume (in mL) = moles × 1000 / concentration
= (0.1133 mol) × (1000 mL) / (4.43 mol/L)
= 25.6 mL
So, the volume of 4.43 M aqueous FeCl3 produced is approximately 25.6 mL.
Based on this calculation, it seems that your initial result of approximately 39,089 mL is incorrect. The correct answer should be 25.6 mL.