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. How much MnO2(s) should be added to excess HCl(aq) to obtain 185 mL of Cl2(g) at 25 °C and 715 Torr? Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq), as described by the chemical equation. How much MnO2(s) should be added to excess HCl(aq) to obtain 185 mL of Cl2(g) at 25 °C and 715 Torr? @Chemistry

Answer 1

Use PV=nRT to figure out how many moles of Cl2 you have. Also convert pressure into either atm (715 Torr = 0.941 atm) and temperature into Kelvin (25C = 298K).

(0.941)(0.185)=n(0.08206)(298) --> n = 0.00712 mol Cl2 produced.

The balanced chemical equation for the reaction is:
MnO2 + 4 HCl --> MnCl2 + Cl2 + 2 H2O

Since Cl2 and MnO2 have the same coefficient, 0.00712 moles of MnO2 were initially reacted. Multiply by molar mass of MnO2 to get mass: 0.00712 * 87 = 0.619 g.