Question

34. When 12 g of methanol (CH3OH) was treated with excess oxidizing agent (MO) 14 g of formic acid (HCOOH) was obtained. Using the following chemical equation, calculate the percent yield. (The reaction is much more complex than this: please ignore the fact that the charges do not balance.) 3CH3OH + 4MnO4 → 3HCOOH + 4MnO2 A. 100% B. 92% C. 81% 79%

Answer 1

- Here the given equation is- 3 CH ₂ OH + 4. MnO4 - 3 HCOOH + 4 Mno, Molar mass of CH₂ OH = Totland 16+7 glmol = 32 g/mol . mass of 3 mol of CH3OH = 3 mol x 22.9.mot! = 96 g Again, Molar mass of HCOOH +1+2+16+16 +179) mol = 46 g/mol mass of 3 mol of it cool = 3x46 g = 138 g So, according to the equation - above chemical

from 96 g of CH₂OH we get 138 g of Aloh " 1.-. . . 138 .n. 96 12 . . . . 138*12 got HODH 96 gor hadh = 17.25 g of HOOH Thus, Theoretical yield = 17.25 g Again, it was given that from 12 g of CH₂OH we obtained 14 g н сън. in Actual yield = 14 g. " % of gield - Actual yield 19 x 100% Theoretical yield 14 x 100 10 = 17.25 = 81.16% 7 81 % So, the correct option is c. 81% Ane