Question

when 12.0 g of methanol (CH3OH) was treated with excess MnO4-, 14.0 g of formic acid (HCOOH) was obtained. using the following chemical equation calculate the percent yield. 3CH3OH + 4MnO4- -> 3HCOOH + 4MnO2

Answer 1

Balanced equation is

3CH₃OH + 4MnO₄^- -----> 3 HCOOH + 4MnO2

number of moles of CH3OH = 12.0g / 32.0419 g/mol = 0.375 mole

from the balanced equation we can say that

3 mole of CH3OH produces 3 mole of HCOOH so

0.375 mole of CH3OH will produce

= 0.375 mole of CH3OH *(3 mole of HCOOH / 3 mole of CH3OH)

= 0.375 mole of HCOOH

1 mole of HCOOH = 46.0254 g

0.375 mole of HCOOH = 17.3 g

Therefore, theoretical yield of HCOOH = 17.3 g

percent yield = (actual yield / theoretical yield)*100

percent yield = (14.0/17.3)*100 = 80.9 %

Therefore, the percent yield = 80.9%