H2C2O4.H2O + 2NaOH ==> Na2C2O4 + 3H2O
mols H2C2O4.H2O = grams/molar mass
mols NaOH = twice that (look at the coefficients in the balanced equation).
Then M NaOH = mols NaOH/L NaOH.
H2C2O4.2H2O(s) is primary standard substance. 2.3688g of oxalic acid hydrate were completely neutralized by 42.56ml of NaOH solution. Calculate the molar concentration of the NaOH solution. Write the balanced equation for the reaction.
2 answers
Oxalic acid dihydrate:
H2C2O4.2H2O(s) same as H6C2O6
H6C2O6(s) + 2NaOH(aq)--> 4H2O + Na2C2O4(aq)
Moles of NaOH:
(2.3688g H6C2O6)*
(1mol H6C2O6/126.064g H6C2O6)*(2mol NaOH/1 mol H6C2O6)
=.037581mol NaOH
Liters of NaOH:
42.56ml= .004256L
NaOH Molarity(molar concentration)
.037581mol NaOH/.004256L NaOH
=8.830M
H2C2O4.2H2O(s) same as H6C2O6
H6C2O6(s) + 2NaOH(aq)--> 4H2O + Na2C2O4(aq)
Moles of NaOH:
(2.3688g H6C2O6)*
(1mol H6C2O6/126.064g H6C2O6)*(2mol NaOH/1 mol H6C2O6)
=.037581mol NaOH
Liters of NaOH:
42.56ml= .004256L
NaOH Molarity(molar concentration)
.037581mol NaOH/.004256L NaOH
=8.830M
2 answers