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) ¡æ 3H2S(g) + 2FeCl3(aq)
Molar Mass (g/mol)
HCl 36.461
FeCl3 162.21
Molar Volume (L)
22.4 at STP
Gas Constant
(L.atm.mol-1.K-1)
0.0821
To find the volume of the 4.43 M aqueous FeCl3 produced, we need to use the stoichiometry of the balanced equation and the concentration of the HCl solution.
First, let's determine the number of moles of HCl in the 120 mL of the 2.8333 M solution using the formula:
moles = volume (L) x concentration (M)
Converting the volume from mL to L:
120 mL = 120 mL x 1 L/1000 mL = 0.12 L
Now, calculating the number of moles of HCl:
moles HCl = 0.12 L x 2.8333 M = 0.339996 moles
According to the balanced equation, 1 mole of HCl reacts with 2 moles of FeCl3, so the number of moles of FeCl3 produced will be half the number of moles of HCl.
moles FeCl3 = 0.339996 moles x 1/2 = 0.169998 moles
Now, let's convert the moles of FeCl3 to volume using the concentration of the FeCl3 solution.
moles = volume (L) x concentration (M)
volume (L) = moles / concentration (M)
volume (L) = 0.169998 moles / 4.43 M = 0.038374 L
Finally, let's convert the volume from L to mL:
volume (mL) = 0.038374 L x 1000 mL/1 L = 38.374 mL
Therefore, approximately 38.374 mL of the 4.43 M aqueous FeCl3 solution is produced.