A 1.575 g sample of ethanedioic acid crystals, H2C2O4. xH20 was dissolved in water and made up to 250cm^3 . One mole of the acid reacts with two moles of NaOH. In a titration, 25.0 cm^3 of this solution of acid reacted with exactly 15.6 cm^3 of 0.160 mol/ dm^3 NaOH. Calculate the value of x
4 answers
What exactly is x?
H2C2O4. x H20
H2C2O4 + 2NaOH ==> Na2CO2O4 + 2H2O
1.575 g H2C2O4.xH2O in 250 mL and 25 cc of that makes the sample titrated 1.575 x (25/250) = 0.1575grams.
mols NaOH = M x L = approx 0.0025 but you need to redo all of this with better numbers.
Convert mols NaOH to mols H2C2O4.xH2O. 1/2 * 0.0025 = approx 0.0012
mols H2C2O4.xH2O = grams/molar mass or
molar mass = g/mol = approx 0.16/0.00125 = approx 128 for H2C2O4.xH2O. You know H2C2O4 is 90 so xH2O must be approx 128-90 = about 38.
38/molar mass H2O = 38/18 = about 2 when rounded to a whole number so x must be 2.
If you redo the problem without the estimates you will get x = 2.
1.575 g H2C2O4.xH2O in 250 mL and 25 cc of that makes the sample titrated 1.575 x (25/250) = 0.1575grams.
mols NaOH = M x L = approx 0.0025 but you need to redo all of this with better numbers.
Convert mols NaOH to mols H2C2O4.xH2O. 1/2 * 0.0025 = approx 0.0012
mols H2C2O4.xH2O = grams/molar mass or
molar mass = g/mol = approx 0.16/0.00125 = approx 128 for H2C2O4.xH2O. You know H2C2O4 is 90 so xH2O must be approx 128-90 = about 38.
38/molar mass H2O = 38/18 = about 2 when rounded to a whole number so x must be 2.
If you redo the problem without the estimates you will get x = 2.
What is the % of carbon in ethanedioc acid crystals
C2H2O4.2H2O
C2H2O4.2H2O
1 answer