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:
[tex]Na_2C_2O_4 + 2HCl = 2NaCl + H_2 + 2CO_2[/tex]
Stoichiometrically, 1 mole of [tex]Na_2C_2O_4[/tex] reacts with 2 moles of [tex]HCl[/tex] for a complete neutralization reaction.
Recall that: mole = [tex]\frac{mass}{molar mass}[/tex]
Mole of 0.550 g sodium oxalate = 0.550/134 = 0.0041 mole
If 1 mole [tex]Na_2C_2O_4[/tex] 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
