How many grams of oxalic acid dihydrate H2C2O4 ·2H2O, a diprotic acid, are needed to react with 20.00 mL of 0.4500 M NaOH?

2 answers:

6 0

With the given components above, the chemical reaction required to solve the problem is H2C2O4*2H2O + 2 NaOH = 4 H2O + Na2C2O4 where 1 mole of diprotic acid needs 2 moles of NaOH to complete the reaction.In this case, given 0.4500 M NaoH and 0.02 L of it, the moles diprotic acid needed is 0.0045 moles. This is equivalent to 0.351 grams.

inna [77]3 years ago

5 0

Answer:

$0.567gC_2H_2O_4.2H_2O$

Explanation:

Hello,

The carried out chemical reaction is:

$HOOCCOOH+2NaOH-->NaOOCCOONa+2H_2O$

As long as the concentration of sodium hydroxide allows us to determine the reacting moles, we perform it as follows:

$n_{NaOH}=0.4500\frac{molNaOH}{L}*0.02000L=0.009000molNaOH$

Now, we can apply, the mole-mass relationship based on the chemical reaction:

$m_{C_2H_2O_4.2H_2O}=0.009000molNaOH*\frac{1molC_2H_2O_4}{2molNaOH}*\frac{1molC_2H_2O_4.2H_2O}{1molC_2H_2O_4} *\frac{126g1molC_2H_2O_4.2H_2O}{1gC_2H_2O_4.2H_2O}\\m_{C_2H_2O_4.2H_2O}=0.567gC_2H_2O_4.2H_2O$