Question

g What is the molarity of hydrochloric acid if 40.95 mL of HCl is required to neutralize 0.550 g of sodium oxalate, Na2C2O4

Answer 1

Answer:

0.0002 M

Explanation:

The molarity of the HCl required would be 0.0002 M.

First, let us consider the balanced equation of the reaction:

$Na_2C_2O_4 + 2HCl = 2NaCl + H_2 + 2CO_2$

Stoichiometrically, 1 mole of $Na_2C_2O_4$ reacts with 2 moles of $HCl$ for a complete neutralization reaction.

Recall that: mole = $\frac{mass}{molar mass}$

Mole of 0.550 g sodium oxalate = 0.550/134 = 0.0041 mole

If 1 mole $Na_2C_2O_4$ requires 2 moles HCl, then 0.0041 mole will require:

    0.0041 x 2 = 0.0082 mole HCl

Volume of the HCl = 40.95 L

Molarity = mole/volume

Hence, molarity of the HCl = 0.0082/40.95 = 0.0002 M