Question

A contains 1.60g of trioxonitrate(v) acid in 250cm³ of the solution.
B contains 9.0gdm3 of XHCO3,
25cm³ portions of B required an average of 24.90cm³ of A for complete neutralization, calculate the:
a) concentration of acid in A in mol/dm³.
b) concentration of XHCO3 in B in mol/dm³.
c) molar mass of XHCO3.
d) value of X.
Equation of the reaction
HNO3+XHCO3- XNO3+CO2+H2O
(H=1, C=12, O=16)

Answer 1

First you should learn a correct name for HNO3. The International Union of Pure and Applied Chemistry (IUPAC) does NOT recognize trioxonitrate(v) acid as a correct name. They do recognize nitric acid as correct. They also recognize hydroxidodioxidonitrogen(v) as correct but most people don't use that name for obvious reasons.
Note that I will use 1 dm^3 = 1 L and the definition of molarity is M = moles/L. Your teacher may prefer you to write M = moles/dm^3 and you should bow to your teacher's wishes. Finally I assume you made a typo and B contains 9.0 g/dm3.
a. (A) = 1.60 g/0.25 L HNO3. moles = 1.60/63 = 0.0254 and M = 0.0254/0.25 L = 0.1016 M HNO3
b. HNO3 + XHCO3 ---> XNO3 + CO2 + H2O
moles HNO3 = M x L = 0.1016 M x 0.02490 = 0.002530
From the equation (1 mol A = 1 mol B) so mols B = 0.002530
molarity of B is mols/L = 0.00253 moles/0.025 L = 0.1012 M = 0.1012 moles/dm3
c. B contains 9.0 g/L and the 9.0 g is 0.1012 mols. Since mole = grams/molar mass or 0.1012 mols = 9.0/molar mass so molar mass = 9.0/0.1012 = about 88.9
HCO3 is about 1 + 12 + 48 = 62 and 89-62 = 27. I don't see anything in column 1 of the periodic table that corresponds to an atomic mass of 27. Obviously the problem was meant to be either Na at 23 or K at 39. Check your numbers. Something is wrong. Either the numbers in the problem are incorrect or you've made another typo.