Asked by alex
A standard sodium carbonate solution is prepared in a 250 cm^3 volumetric flask
during a titration,20 cm^3 of a 0,1 mol.dm^-3 nitric acid solution neutralises 25 cm^3 of the above standard solution according to the following balanced equation
2HNO3(aq) Na2CO3(aq)=>2NaNO3(aq) H20(l) CO2(g)
calculate the mass of sodium carbonate used to prepare the standard solution in the volumetric flask
during a titration,20 cm^3 of a 0,1 mol.dm^-3 nitric acid solution neutralises 25 cm^3 of the above standard solution according to the following balanced equation
2HNO3(aq) Na2CO3(aq)=>2NaNO3(aq) H20(l) CO2(g)
calculate the mass of sodium carbonate used to prepare the standard solution in the volumetric flask
Answers
Answered by
DrBob222
2HNO3(aq) Na2CO3(aq)=>2NaNO3(aq) H20(l) CO2(g)
mols HNO3 = M x L = ?
Using the coefficients in the balanced equation, convert mols HNO3 to mols Na2CO3 (That will be 1/2 mols HNO3)
Then grams Na2CO3 = mols Na2CO3 x molar mass Na2CO3. This will be the grams in the 25 cc of the solution. That x (250/25) gives the grams in the 250 cc solution.
mols HNO3 = M x L = ?
Using the coefficients in the balanced equation, convert mols HNO3 to mols Na2CO3 (That will be 1/2 mols HNO3)
Then grams Na2CO3 = mols Na2CO3 x molar mass Na2CO3. This will be the grams in the 25 cc of the solution. That x (250/25) gives the grams in the 250 cc solution.
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