# 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?

Feb 21, 2015

This reaction will produce $\text{2.63 g}$ of $C {l}_{2}$ gas.

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

$M n {O}_{2 \left(s\right)} + 4 H C {l}_{\left(a q\right)} \to C {l}_{2 \left(g\right)} + M n C {l}_{2 \left(a q\right)} + 2 {H}_{2} {O}_{\left(l\right)}$

Now look at the mole ratios you have between $M n {O}_{2}$, $H C l$, and $C {l}_{2}$: 1 mole of $M n {O}_{2}$ needs 4 moles of $H C l$ in order to produce 1 mole of $C {l}_{2}$ gas.

SInce you can calculate how many moles of both $M n {O}_{2}$ and $H C l$ 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.

$\text{5.00 g" * ("1 mole")/("87.0 g") = "0.0575 moles}$ $M n {O}_{2}$

Use the molarity of the $H C l$ solution to determine how many moles you have

$C = \frac{n}{V} \implies {n}_{H C l} = C \cdot V = \text{6.0 M" * 25 * 10^(-3)"L" = "0.15 moles HCl}$

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

$\text{0.0575 moles MnO"_2 * ("4 moles HCl")/("1 mole MnO"_2) = "0.0230 moles HCl}$

This means that $H C l$ is acting as a limiting reagent and will determine how much $M n {O}_{2}$ actually reacts

$\text{0.15 moles HCl" * ("1 mole MnO"_2)/("4 moles HCl") = "0.0375 moles}$ $M n {O}_{2}$

This will be equal to the number of $C {l}_{2}$ moles produced - remember the $\text{1:1}$ mole ratio between the two compounds. Therefore, the mass of $C {l}_{2}$ gas will be

$\text{0.0375 moles" * ("70.0 g")/("1 mole") = "2.63 g}$

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

$M n {O}_{2 \left(s\right)} + 4 {H}_{\left(a q\right)}^{+} + 2 C {l}_{\left(a q\right)}^{-} \to C {l}_{2 \left(g\right)} + M {n}_{\left(a q\right)}^{2 +} + 2 {H}_{2} {O}_{\left(l\right)}$