Gas Law chemistry

Question

Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq), as described by the chemical equation

MnO^2(s)+4HCl(aq)⟶MnCl2(aq)+2H^2O(l)+Cl^2(g)

How much MnO^2(s) should be added to excess HCl(aq) to obtain 195 mL Cl2(g) at 25 °C and 735 Torr?

Answer 1

You shouldn't use MnO^2 and Cl^2 to designate MnO₂ and Cl₂ because the ^2 suggests a superscript, not a subscript. Now, on to the problem at hand.

MnO₂(s) + 4HCl(aq) ==> MnCl₂(aq) + 2H₂O(l) + Cl₂(g)

Use the ideal gas law to find moles of Cl₂ being produced. Then use stoichiometry and dimensional analysis to find moles and mass of MnO₂ needed.

PV = nRT and solve for n (moles)

P = pressure = 735 torr x 1 atm/760 torr = 0.967 atm

V = volume = 195 ml = 0.195 L

n = moles = ?

R = gas constant = 0.0821 Latm/Kmol

T = temp in K = 25C + 273 = 298K

n = PV/RT = (0.967)(0.195) / (0.0821)(298)

n = 0.003755 moles Cl₂(g)

moles MnO₂ needed = 0.003755 mol Cl₂ x 1 mol MnO₂ / mol Cl₂ = 0.003755 moles MnO₂ needed

mass MnO₂ needed = 0.003755 moles x 86.94 g/mol = 0.326 g MnO₂ needed