USA: +1 (725) 712-1730 
Sign in
What volume of chlorine gas at 25 degree C and 0.950 atm can be produced by the reaction of 12.0 g of MnO_2 in excess HCl(aq)? MnO_2(s) + 4HCl(aq) rightarrow MnCl_2(aq) + 2H_2O(l) + Cl_2(g) (R = 0.08206 L middot atm/k middot mol) 5.36 times 10^-3 L 0.138 L 0.282 L 3.09 L 3.55 L
Expert Answer
Answer
MnO2 (g) + 4HCl (aq) ————-> MnCl2 (aq) + Cl2(g) + 2H2O(l)
1 mole of MnO2 react with HCl to gives 1 mole of Cl2
86.94 of MnO2 react with HCl to gives 1 mole of Cl2
12g of MnO2 react with HCl to gives = 1*12/86.94 = 0.138 moles of Cl2
PV = nRT
n = 0.138 moles
P = 0.95atm
T = 250C = 25 + 273 = 298K
R = 0.0821L-atm/mole-K
PV = nRT
V = nRT/P
= 0.138*0.0821*298/0.95
= 3.55L >>>>>>answer
volume of Cl2 is 3.55L
