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)
a) The molar mass of trioxonitrate(V) acid (HNO3) is 1(1) + 1(14) + 3(16) = 63 g/mol.
To find the concentration of acid in A in mol/dm³, we need to find the number of moles of HNO3 in 1.60 g.
Number of moles of HNO3 = Mass / Molar mass
Number of moles of HNO3 = 1.60 g / 63 g/mol
Since the volume of A is given in cm³, we need to convert it to dm³.
1 dm³ = 1000 cm³
250 cm³ = 250/1000 dm³ = 0.25 dm³
Concentration of acid in A = Number of moles of HNO3 / Volume of A in dm³
Concentration of acid in A = (1.60 g / 63 g/mol) / 0.25 dm³
Concentration of acid in A = 0.0254 mol/dm³
b) The concentration of XHCO3 in B is given as 9.0 g/dm³.
To find the concentration of XHCO3 in mol/dm³, we need to find the number of moles of XHCO3 in 9.0 g.
Number of moles of XHCO3 = Mass / Molar mass
Number of moles of XHCO3 = 9.0 g / Molar mass of XHCO3
c) To find the molar mass of XHCO3, we need to know the value of X, which is not provided in the question. Could you please provide the value of X?
d) The value of X cannot be determined without knowing the molar mass of XHCO3 and the balanced equation.
To solve this problem, we need to calculate the concentration of acid in solution A and the concentration of XHCO3 in solution B. Let's go step by step.
Step 1: Calculate the number of moles of trioxonitrate(V) acid in solution A.
Given:
Mass of acid in A = 1.60 g
Volume of solution A = 250 cm³
To calculate the number of moles, we need to use the formula:
Moles = Mass / Molar Mass
The molar mass of trioxonitrate(V) acid (HNO3) is:
H = 1 (1 atom) + N = 14 (1 atom) + O = 16 (3 atoms) = 1 + 14 + 48 = 63 g/mol.
Using the formula:
Moles = 1.60 g / 63 g/mol = 0.0254 mol
Therefore, the number of moles of trioxonitrate(V) acid in solution A is 0.0254 mol.
Step 2: Calculate the concentration of acid in solution A in mol/dm³.
Given:
Volume of solution A = 250 cm³ = 0.25 dm³
To calculate the concentration, we use the formula:
Concentration (molarity) = Moles / Volume
Concentration = 0.0254 mol / 0.25 dm³ = 0.1016 mol/dm³
Therefore, the concentration of trioxonitrate(V) acid in solution A is 0.1016 mol/dm³.
Step 3: Calculate the concentration of XHCO3 in solution B in mol/dm³.
Given:
Mass of XHCO3 in B = 9.0 g/dm³
Volume of solution B = 25 cm³ = 0.025 dm³
To calculate the concentration, we use the formula:
Concentration (molarity) = Moles / Volume
First, we need to calculate the moles of XHCO3 in solution B using the molar mass.
The molar mass of XHCO3 (X = unknown) is:
H = 1 (1 atom) + C = 12 (1 atom) + O = 16 (3 atoms) = 1 + 12 + 48 = 61 g/mol.
Moles = 9.0 g / 61 g/mol = 0.1475 mol
Concentration = 0.1475 mol / 0.025 dm³ = 5.9 mol/dm³
Therefore, the concentration of XHCO3 in solution B is 5.9 mol/dm³.
Step 4: Calculate the molar mass of XHCO3.
Given:
H = 1, C = 12, O = 16.
Using the molar mass formula:
Molar Mass = H + C + (O * 3) = 1 + 12 + (16 * 3) = 1 + 12 + 48 = 61 g/mol.
Therefore, the molar mass of XHCO3 is 61 g/mol.
Step 5: Calculate the value of X in the equation.
The equation of the reaction is:
HNO3 + XHCO3 -> XNO3 + CO2 + H2O
From the equation, we can determine the stoichiometry, which shows that 1 mole of HNO3 reacts with 1 mole of XHCO3 to produce 1 mole of XNO3.
Since the average volume of A required for neutralization is 24.90 cm³, which is the same as the volume of XNO3, we can assume that X is 1.
Therefore, the value of X is 1.
To summarize the results:
a) The concentration of acid in A is 0.1016 mol/dm³.
b) The concentration of XHCO3 in B is 5.9 mol/dm³.
c) The molar mass of XHCO3 is 61 g/mol.
d) The value of X is 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.
To solve this problem, we'll need to follow a step-by-step process. Let's break it down into smaller parts:
a) Calculation of concentration of acid in A in mol/dm³:
First, we need to calculate the number of moles of trioxonitrate(V) acid in 1.60g. To do this, we divide the mass by the molar mass of HNO3.
Molar mass of HNO3 (H=1, N=14, O=16):
= (1*1) + (1*14) + (3*16)
= 1 + 14 + 48
= 63 g/mol
Number of moles of HNO3 in 1.60g:
= mass / molar mass
= 1.60g / 63 g/mol
Now, we need to convert the volume from cm³ to dm³.
1 dm³ = 1000 cm³
So, 250 cm³ = 250/1000 dm³ = 0.25 dm³
Concentration of acid in A in mol/dm³:
= moles / volume
= (1.60g / 63 g/mol) / 0.25 dm³
= 0.040 mol/dm³
b) Calculation of concentration of XHCO3 in B in mol/dm³:
The concentration provided is given in g/dm³, so we need to convert it to mol/dm³.
To convert from grams to moles, we divide the mass by the molar mass of XHCO3.
The molar mass of XHCO3 can be calculated as follows:
Molar mass of XHCO3 (X=atomic mass of X, H=1, C=12, O=16):
= atomic mass of X + (1*1) + (1*12) + (3*16)
Since the atomic mass of X is not given, we'll have to proceed with this information.
c) Calculation of molar mass of XHCO3:
The molar mass of XHCO3 cannot be determined without knowing the atomic mass of X. If you have additional information about the nature or atomic mass of X, please provide it, and we can proceed with the calculation.
d) Calculation of the value of X:
The equation of the reaction given is:
HNO3 + XHCO3 -> XNO3 + CO2 + H2O
From this equation, we can see that 1 mole of HNO3 reacts with 1 mole of XHCO3 to produce 1 mole of XNO3.
Since the average volume of A required to neutralize each 25 cm³ portion of B is 24.90 cm³, we can assume that 1 mole of HNO3 reacts with 1 mole of XHCO3.
Therefore, the value of X is 1.