68 363
Assignments Done
98,7%
Successfully Done
In November 2018

Answer to Question #57245 in General Chemistry for Delaney

Question #57245
When 120 grams of glucose (C6H12O6) is burned in the lab 65 grams of water is produced. What is the percent yield of water?
Expert's answer
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 %

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be first!

Leave a comment

Ask Your question

Submit
Privacy policy Terms and Conditions