AHan 86212 7. Ethanol burns in excess oxygen to form CO(B) and H,O(B) according

Question

AHan 86212 7. Ethanol burns in excess oxygen to form CO(B) and H,O(B) according to this balanced equation. ener9f CMOH(g) + 30,Ig) 2co,(g) + 3H,0 g) What value is closest to the volume of CO(g). measured at 200K and 1 atm, produced from the combustion of 0.25 mol of C,H,OH(e) N DST 4.IL 8. SH,GOdaq) + 2MnO. (aq) + 6H-(aq) 2Mn).(aq) + 10cole) + 8H,O(1) Oxalic acid, H,C,O, reacts with permanganate ion according to the balanced equation above. How many mL of 0.0154 M KMnO, solution are required to react with 25.0 mL of 0.0208 M H,C,O solution? Reference Sheet

Explanation / Answer

7.

C2H5OH(g) + 3O2(g) ------------> 2Co2(g) + 3H2O(g)
1 moles of C2H5OH on combustion to gives 2 moles of CO2
0.25moles of C2H5OH on combustion to gives = 2*0.25/1 = 0.5 moles of CO2
n = 0.5moles
T = 200K
P   = 1atm
PV = nRT
V = nRT/P
   = 0.5*0.0821*200/1   = 8.21L
volume of Co2 is 9.21L

8. 5H2C2O4 (aq) + 2MnO4^- (aq) + 6H^+ (aq) ----------> Mn^2+ (aq) + 10CO2(g) + 8H2O(l)
   5 moles        2 moles

KMnO4                                           H2C2O4
M1 = 0.0154M                                   M2 = 0.0208M
V1 =                                           V2 = 25ml
n1 = 2                                          n2 = 5
             M1V1/n1   =    M2V2/n2
                V1     =    M2V2n1/M1n2
                       =   0.0208*25*2/0.0154*5
                       = 13.5ml
The volume of KMnO4 is 13.5ml

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