Question

50ml of 6.0M NaOH provides an excess of NaOH for the reaction. How many mLs of 6.0M NaOh are actually needed to react with 10.0g of triglyceride?​

Answer 1

The volume of 6 M NaOH is 5.5 ml

Step-by-step explanation:

The equation of reaction is

C₃H₅(C₁₇H₃₅COO)₃+3 NaOH → C₃H₅(OH)₃+3 C₁₇H₃₅COONa

The molar mass of triglyceride M is 890 g/mole, so in 10.0 g of its are

`(10)/(890)=0.011 mole`

From the equation of reaction one mole of triglyceride reacts with three moles of NaOH.

So,

the quantity moles of NaOH which actually needed to react is:

0.011 x 3=0.033 moles.

These quantity moles of NaOH are consist in
`(0.033)/(6)X1000=5.5 ml`

∴The volume of 6 M NaOH is 5.5 ml