A solution of trioxonitrate(v)acid contains 0.63g in 100cm³. 28.0cm³of this solution neutralized 25cm³of sodiumtrioxocarbonate(iv) solution. Calculate the molar concentration and mass concentration of sodiumtrioxocarbonate(iv)
HNO3 = 0.63 g/100 c
(HNO3) = mols/L = g/molar mass/L = 0.63/63/0.100L = 0.1 M
2HNO3 + Na2CO3 ==> 2NaNO3 + H2O + CO2
millimols HNO3 = mL x M = 28.0 mL x 0.1 M = 2.80
2.8 mmols HNO3 will neutralize 2.80 x (1 mol Na2CO3/2 mol HNO3) = 1.40 mmols Na2CO3 .
The (Na2CO3) = mols/L = millimoles/mL = 14.0/25 = ? M
To calculate the molar concentration and mass concentration of sodium trioxocarbonate(IV) (Na2CO3) solution, we need to use the equation of the reaction between trioxonitrate(V) acid (HNO3) and sodium trioxocarbonate(IV) (Na2CO3):
2HNO3 + Na2CO3 -> 2NaNO3 + CO2 + H2O
Given information:
- Mass of HNO3 in the solution = 0.63 g
- Volume of the solution = 100 cm³
- Volume of HNO3 solution used for neutralization = 28.0 cm³
- Volume of Na2CO3 solution used for neutralization = 25.0 cm³
Step 1: Calculate the number of moles of HNO3 used for neutralization.
Molar mass of HNO3 = 1(Atomic mass of H) + 14(Atomic mass of N) + 48(Atomic mass of O) = 63 g/mol
Number of moles of HNO3 = Mass of HNO3 / Molar mass of HNO3 = 0.63 g / 63 g/mol = 0.01 mol
Step 2: Use the stoichiometry of the reaction to determine the number of moles of Na2CO3 used for neutralization.
From the balanced equation, we can see that the ratio of moles of HNO3 to Na2CO3 is 2:1.
Number of moles of Na2CO3 = 0.01 mol / 2 = 0.005 mol
Step 3: Calculate the molar concentration of Na2CO3.
Molar concentration of Na2CO3 = Number of moles / Volume in liters
Volume in liters = 25.0 cm³ / 1000 = 0.025 L
Molar concentration of Na2CO3 = 0.005 mol / 0.025 L = 0.2 mol/L
Step 4: Calculate the mass concentration of Na2CO3.
Mass concentration of Na2CO3 = Mass of Na2CO3 / Volume in liters
Mass of Na2CO3 = Molar mass of Na2CO3 = 2(Atomic mass of Na) + Atomic mass of C + 3(Atomic mass of O) = 46 g/mol
Mass concentration of Na2CO3 = (0.2 mol/L) * (46 g/mol) = 9.2 g/L
Therefore, the molar concentration of Na2CO3 is 0.2 mol/L, and the mass concentration of Na2CO3 is 9.2 g/L.
To calculate the molar concentration and mass concentration of sodium trioxocarbonate(IV) solution, we first need to determine the amount of trioxonitrate(V) acid that was neutralized during the reaction.
From the given information, we know that 28.0 cm³ of the trioxonitrate(V) acid solution neutralized 25 cm³ of the sodium trioxocarbonate(IV) solution. Since the two substances react in a 1:1 ratio, the number of moles of trioxonitrate(V) acid can be determined as follows:
Moles of trioxonitrate(V) acid = Volume of trioxonitrate(V) acid solution × Molar concentration of trioxonitrate(V) acid
The volume of trioxonitrate(V) acid solution is given as 28.0 cm³, and we need to find the molar concentration of the solution.
Molar concentration = Mass concentration ÷ Molar mass
Let's start by calculating the moles of trioxonitrate(V) acid:
Moles of trioxonitrate(V) acid = (Mass of trioxonitrate(V) acid ÷ Volume of trioxonitrate(V) acid) × 1000
Given that the mass of trioxonitrate(V) acid is 0.63 g and the volume of the solution is 100 cm³, we can substitute these values into the equation:
Moles of trioxonitrate(V) acid = (0.63 g ÷ 100 cm³) × 1000 = 6.3 mol/dm³
Now we can use the 1:1 ratio to determine the moles of sodium trioxocarbonate(IV) solution:
Moles of sodium trioxocarbonate(IV) = Moles of trioxonitrate(V) acid = 6.3 mol/dm³
Finally, we can calculate the molar concentration and mass concentration of sodium trioxocarbonate(IV) solution:
Molar concentration of sodium trioxocarbonate(IV) = Moles of sodium trioxocarbonate(IV) ÷ Volume of sodium trioxocarbonate(IV) solution
Given that the volume of sodium trioxocarbonate(IV) solution is 25 cm³:
Molar concentration of sodium trioxocarbonate(IV) = 6.3 mol/dm³ ÷ 25 cm³ = 0.252 mol/dm³
Mass concentration of sodium trioxocarbonate(IV) = Mass of sodium trioxocarbonate(IV) ÷ Volume of sodium trioxocarbonate(IV) solution
To calculate the mass concentration, we need the molar mass of sodium trioxocarbonate(IV) (Na2CO3). The molar mass is 22.99 g/mol (sodium) + 12.01 g/mol (carbon) + (3 × 16.00 g/mol) (oxygen).
Mass concentration of sodium trioxocarbonate(IV) = (0.252 mol/dm³ × molar mass of Na2CO3) ÷ 25 cm³
Substituting the molar mass, we get:
Mass concentration of sodium trioxocarbonate(IV) = (0.252 mol/dm³ × 106.0 g/mol) ÷ 25 cm³ = 1.06 g/dm³
Therefore, the molar concentration of sodium trioxocarbonate(IV) solution is 0.252 mol/dm³, and the mass concentration is 1.06 g/dm³.