Question
C7H6O3+CH3OH→C8H8O3+H2O
In any experiment, 1.50g of salicyclic acid is reacted with 11.20g of methanol. The yield of methyl salicylate, C8H8O3, is 1.31g. What is the percent yield?
In any experiment, 1.50g of salicyclic acid is reacted with 11.20g of methanol. The yield of methyl salicylate, C8H8O3, is 1.31g. What is the percent yield?
Answers
DrBob222
1. Writ and balance the equation.
2. Convert 11.2 g methanol to moles. moles = grams/molar mass.
3. Using the coefficients in the balanced equation, convert moles methanol to moles of the product.
4. Now convert moles of the product to grams. g = moles x molar mass. This is the theoretical yield.
5. %yield = (actual yield/theoretical yield)*100 = ??
2. Convert 11.2 g methanol to moles. moles = grams/molar mass.
3. Using the coefficients in the balanced equation, convert moles methanol to moles of the product.
4. Now convert moles of the product to grams. g = moles x molar mass. This is the theoretical yield.
5. %yield = (actual yield/theoretical yield)*100 = ??
Molly
From my understanding the equation is already balanced, correct?
My answer: 26.7%, is this correct?
Work: (11.20gCH3OH)(1 mol CH3OH)/32.05gCH3OH=0.3495gCH3OH
0.3495g/1.31g= 26.68%
My answer: 26.7%, is this correct?
Work: (11.20gCH3OH)(1 mol CH3OH)/32.05gCH3OH=0.3495gCH3OH
0.3495g/1.31g= 26.68%