How much MnO2(s) should be added to excess HCl(aq) to obtain 195 mL Cl2(g) at 25 °C and 715 Torr g

THIS IS THE COMPLETE 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?

0.62901mol of MnO2(s) should be added

Explanation:

Given:

P = 715/760 = 0.94078atm

v=195ml=0.195l

n = ? moles have to find

R = 0.0821 L atm/K/mole

T = 25 + 273 = 298 K

Then we will make use of below formula

PV = nRT

Insert the values

0.94078*0.195=n 0.0821*298

24.466n=0.1740443

n=0.174/24.466

n=0.007235 nb of moles of cl2

as 1 mole of Cl2 were obtained from 1 mole of MnO2

so 0.007235 of chlorine must have come from

0.007235 moles of MnO2

1 mole of MnO2 = 86.94 g/mole

so 0.007235 moles of MnO2== 86.94* 0.007235

=0.62901

there are two significant figures.