If 5.00g of MnO2 reacts w/ 25.0mL of 6M HCL calculate the mass of Cl2 produced? I figured out MnO2 but stuck with HCl. I did 25mL HCl x(0.001L/1mL)x (6mol HCl/1 L) but I don't know where I should go now?

1 Answer
Feb 21, 2015

This reaction will produce "2.63 g" of Cl_2 gas.

Start with the balanced chemical equation, that is a must for any stoichiometry problem.

MnO_(2(s)) + 4HCl_((aq)) -> Cl_(2(g)) + MnCl_(2(aq)) + 2H_2O_((l))

Now look at the mole ratios you have between MnO_2, HCl, and Cl_2: 1 mole of MnO_2 needs 4 moles of HCl in order to produce 1 mole of Cl_2 gas.

SInce you can calculate how many moles of both MnO_2 and HCl you have, you need to determine whether or not one of them acts as a limiting reagent, i.e. is in insufficient amount when compared with the other one.

"5.00 g" * ("1 mole")/("87.0 g") = "0.0575 moles" MnO_2

Use the molarity of the HCl solution to determine how many moles you have

C = n/V => n_(HCl) = C * V = "6.0 M" * 25 * 10^(-3)"L" = "0.15 moles HCl"

However, according to the mole ratio, you would have needed

"0.0575 moles MnO"_2 * ("4 moles HCl")/("1 mole MnO"_2) = "0.0230 moles HCl"

This means that HCl is acting as a limiting reagent and will determine how much MnO_2 actually reacts

"0.15 moles HCl" * ("1 mole MnO"_2)/("4 moles HCl") = "0.0375 moles" MnO_2

This will be equal to the number of Cl_2 moles produced - remember the "1:1" mole ratio between the two compounds. Therefore, the mass of Cl_2 gas will be

"0.0375 moles" * ("70.0 g")/("1 mole") = "2.63 g"

You can adjust this to the number of sig figs you actually have, since I think 6 M was actually 6.0, or 6.00 M.

SIDE NOTE The net ionic equation for this reaction is

MnO_(2(s)) + 4H_((aq))^(+) + 2Cl_((aq))^(-) -> Cl_(2(g)) + Mn_((aq))^(2+) + 2H_2O_((l))