h o l y
what grade is this?
25cm3 of 3g hydrated trioxocarbonate (iv) solution (Na2Co3. xH20) per 250cm3 was neutralised by 13.20cm3 of 0.16M Hcl solution. write the equation of reaction
2 answers
You should learn to name this compound properly. It is hydrated sodium carbonate.
Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2. That's the equation. You can determine the x in the hydrate if you wish.
millimoles HCl = mL x M = 13.20 mL x 0.16 M = 2.11)rounded)
millimoles Na2CO3.xH2O = 1/2 that or 1.056 millimoles or 0.001056 moles of the compound in the 25 cc aliquot. That came from 250 cc solution so the mols in the original 250 cc is 10 x 0.001056 = 0.01056 and that was for a 3 grams sample.
mols = g/molar mass = 0.01056 = 3.00/molar mass or
molar mass = 3.00/0.01056 = 284 rounded.
Na2CO3 is 2*23 + 12 + 48 = 106 so the water is what's left or
284 - 106 = 186.4. That is 186.4/18.0 = 9.9 which rounds to 10 so the formula for the initial hydrated salt is Na2CO3.10H2O
Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2. That's the equation. You can determine the x in the hydrate if you wish.
millimoles HCl = mL x M = 13.20 mL x 0.16 M = 2.11)rounded)
millimoles Na2CO3.xH2O = 1/2 that or 1.056 millimoles or 0.001056 moles of the compound in the 25 cc aliquot. That came from 250 cc solution so the mols in the original 250 cc is 10 x 0.001056 = 0.01056 and that was for a 3 grams sample.
mols = g/molar mass = 0.01056 = 3.00/molar mass or
molar mass = 3.00/0.01056 = 284 rounded.
Na2CO3 is 2*23 + 12 + 48 = 106 so the water is what's left or
284 - 106 = 186.4. That is 186.4/18.0 = 9.9 which rounds to 10 so the formula for the initial hydrated salt is Na2CO3.10H2O
4 answers