Nitric acid and sodium carbonate are names approved by the IUPAC. In my book, those are the preferred names for those compounds.
..............2HNO3 + Na2CO3 ==> 2NaNO3 + H2O + CO2
mols HNO3 = grams/molar mass = 0.67/63 = 0.0106 in 100 cc. How many moles in the 31.0 cc? That's 0.0106 x (31.0 cc/100 cc) = 0.00330 (rounded).
This is equivalent to how any mols of Na2CO3? That's
0.00330 mols HNO3 x (1 mol Na2CO3/2 mols HNO3) = 0.00165
Then M Na2CO3 = mols/L = 0.00165 mols/0.025 L = ?
A solution of trioxonitrate (V) acid contained 0.67g in 100cm3. 31.0cm3 of this solution
neutralized 25cm3 of a sodium trioxocarbonate (IV) solution. Calculate the concentration of the
trioxocarbonate (IV) solution. (HNO3 = 63, Na2CO3 = 106)
2 answers
Yes
1 answer