Asked by Michael
A solution of trioxonitrate v contains 0•67g in 100cm^3. 31•00cm^3 of this solution neutralize 25cm^3 of Na2CO3 solution. Calculate the concentration of the Na2CO3 solution.
Answers
Answered by
DrBob222
Note: a better IUPAC name, in my opinion, is nitric acid.
What is the concentration of the HNO3.? That is mols/L of solution.
mol = g/molar mass = 0.67/63 = 0.0106
M = mols/L = 0.0106/0.100 L = 0.106 M. You took 31.00 cc of that solution.You took how many millimoles of that solution? That is mL x M = 31 x 0.106 = 3.286 millimols HNO3.
Na2CO3 + 2HNO3 ==> 2NaNO3 + 2H2O + CO2.
3.286 mmols HNO3 x (1 mols Na2CO3/2 mols HNO3) = 3.286/2 = 1.643 millimols Na2CO3. M Na2CO3 = millimoles/mL = 1.643/25 cc = ? M
Post your work if you get stuck.
What is the concentration of the HNO3.? That is mols/L of solution.
mol = g/molar mass = 0.67/63 = 0.0106
M = mols/L = 0.0106/0.100 L = 0.106 M. You took 31.00 cc of that solution.You took how many millimoles of that solution? That is mL x M = 31 x 0.106 = 3.286 millimols HNO3.
Na2CO3 + 2HNO3 ==> 2NaNO3 + 2H2O + CO2.
3.286 mmols HNO3 x (1 mols Na2CO3/2 mols HNO3) = 3.286/2 = 1.643 millimols Na2CO3. M Na2CO3 = millimoles/mL = 1.643/25 cc = ? M
Post your work if you get stuck.
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