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?
Answer:
0.62901mol of MnO2(s) should be added
Step-by-step 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