Answer to Question #57245 in General Chemistry for Delaney

Question

When 120 grams of glucose (C6H12O6) is burned in the lab 65 grams of water is produced. What is the percent yield of water?

Answer 1

Solution
C6H12O6 + 6O2 = 6CO2 + 6H2O
M(C6H12O6) = 6*12 + 1*12 + 6*16 = 180 (g/mol)
n(C6H12O6) = m/M = 120/180 = 0.67 (mol)
n(H2O) = n(C6H12O6) * 6 = 4 (mol)
m(H2O) = n*M = 4*18 = 72 (g)
yield = m’/m * 100% = 65/72 * 100% = 90.27 %

Answer
yield = 90.27 %