Question

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

Answer 1

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

**Solution:**

The equation of reaction is

$$C_3H_5(C_{17}H_{35}COO)_3 + 3NaOH = C_3H_5(OH)_3 + 3C_{17}H_{35}COONa$$

The molar mass of triglyceride M(C₃H₅(C₁₇H₃₅COO)₃) is 890 g/mol, so in 10.0g of its are

$\frac{10.0}{890} = 0.011$ 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: 0.011·3=0.033 moles. These

quantity moles of NaOH are consistent in $\frac{0.033}{6.0 \cdot 1000} = 5.5\mathrm{mL}$

**Answer:**

The volume of 6 M NaOH is 5.5 mL.