Chemistry help please

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

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

How much MnO2(s) should be added to excess HCl(aq) to obtain 145 mL Cl2(g) at 25 °C and 735 Torr?

Answer 1

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

Convert mls Cl2 to moles under the specified conditions. Use the ideal gas law:

PV = nRT

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

V = volume in L = 145 ml x 1 L / 1000 ml = 0.145 L

n = moles = ?

R = gas constant = 00821 Latm/Kmol

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

Solving for moles (n), we have...

n = PV/RT = (0.967)(0.145) / (0.0821)(298)

n = 0.00573 moles

Using the stoichiometry of the balanced equation and dimensional analysis, we can solve for the amount of MnO2 needed.

0.00573 mols Cl₂(g) x 1 mol MnO₂ / 1 mol Cl₂ = 0.00573 mols MnO₂ needed

mass MnO₂ = 0.00573 mols x 86.9 g /mol = 0.498 g MnO₂ needed