How much MnO2(s) should be added to excess HCl(aq) to obtain 135 mL Cl2(g) at 25 °C and 102 kPa ?

Answer 1

First, we MUST write a correctly balanced equation for the reaction taking place...

MnO₂(s) + 4HCl(aq) ==> Cl₂(g) + MnCl₂ + 2H₂O ... balanced equation

Next, we convert mls Cl2 @25º and 102 kPa to moles, and then find moles of MnO2 needed.

moles Cl2:

PV = nRT

P = pressure = 102 kPa x 1 atm 101.325 kPa = 1.00 atm

V = volume in liters = 135 mls x 1 L /1000 mls = 0.135 L

n = moles = ?

R = gas constant = 0.0821 Latm/Kmol

T = temperature in K = 25ºC +273 = 298K

Solving for n (moles of Cl₂)...

n = PV/RT = (1)(0.135) / (0.0821)(298)

n = 0.00552 moles Cl₂

Converting this to moles of MnO₂, we have...

0.00552 mol Cl₂ x 1 mol MnO₂ / mol Cl₂ = 0.00552 mols MnO₂ needed

If you want this value in grams, simply multiply it by the molar mass for MnO₂

The question didn't specify grams or moles.