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)

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 = ?

Yes

Well, let's see if I can calculate this with a funny twist!

To find the concentration of the sodium trioxocarbonate (IV) solution, we'll need to do a little math.

First, we'll start by finding the number of moles of trioxonitrate (V) acid in the 31.0 cm3 of solution.

The molar mass of HNO3 is 63 g/mol. So, converting the mass of 0.67 g to moles, we get:

0.67 g / 63 g/mol = 0.0106 mol

Now, we can use the balanced equation between trioxonitrate (V) acid and sodium trioxocarbonate (IV) to determine the mole ratio:

2HNO3 + Na2CO3 -> 2NaNO3 + H2O + CO2

Based on the equation, we can see that 2 moles of trioxonitrate (V) acid neutralize 1 mole of sodium trioxocarbonate (IV).

Since 31.0 cm3 of the trioxonitrate (V) acid solution neutralized 25 cm3 of the sodium trioxocarbonate (IV) solution, we know that the ratio of the two solutions is 31.0/25.0.

So, the moles of sodium trioxocarbonate (IV) in 31.0 cm3 of solution can be calculated as follows:

0.0106 mol * (25.0 cm3 / 31.0 cm3) = 0.0085 mol

Finally, we can calculate the concentration of the sodium trioxocarbonate (IV) solution by dividing the moles by the volume:

Concentration = 0.0085 mol / (25 cm3 / 1000) = 0.34 mol/dm3

Therefore, the concentration of the sodium trioxocarbonate (IV) solution is 0.34 mol/dm3.

Now, wasn't that a fun way to calculate the concentration?

To calculate the concentration of the sodium trioxocarbonate (IV) solution, we can use a balanced chemical equation for the neutralization reaction between the trioxonitrate (V) acid and the sodium trioxocarbonate (IV) solution:

2HNO3 + Na2CO3 -> 2NaNO3 + H2O + CO2

From the balanced equation, we can see that the molar ratio between HNO3 and Na2CO3 is 2:1.

Given that 31.0cm3 of the trioxonitrate (V) acid solution neutralized 25cm3 of the sodium trioxocarbonate (IV) solution, we can set up a proportion to find the moles of HNO3 and Na2CO3:

0.67g HNO3 / 100cm3 HNO3 = x g Na2CO3 / 25cm3 Na2CO3

Now, we can convert the grams of HNO3 to moles using its molar mass:

0.67g HNO3 / 63g/mol = x g Na2CO3 / 25cm3 Na2CO3

Simplifying further:

0.0106 mol HNO3 = x g Na2CO3 / 25cm3 Na2CO3

Since the molar ratio between HNO3 and Na2CO3 is 2:1, the moles of Na2CO3 are half of the moles of HNO3:

0.0106 mol HNO3 = 0.0053 mol Na2CO3 / 25cm3 Na2CO3

Finally, we can calculate the concentration of the sodium trioxocarbonate (IV) solution by dividing the moles by the volume (in dm3) of the solution:

0.0053 mol Na2CO3 / 0.025 dm3 Na2CO3 = 0.212 mol/dm3 Na2CO3

Therefore, the concentration of the sodium trioxocarbonate (IV) solution is 0.212 mol/dm3.

To calculate the concentration of the sodium trioxocarbonate (IV) solution, we need to use the concept of stoichiometry and the balanced chemical equation for the reaction.

The balanced chemical equation for the reaction between trioxonitrate (V) acid (HNO3) and sodium trioxocarbonate (IV) (Na2CO3) is:

2HNO3 + Na2CO3 -> 2NaNO3 + H2O + CO2

First, let's calculate the number of moles of trioxonitrate (V) acid in 31.0 cm3 of the solution.

Given:
Mass of HNO3 = 0.67 g
Volume of HNO3 solution = 100 cm3

Using the formula:
Number of moles = mass / molar mass

Molar mass of HNO3 = 63 g/mol
Number of moles of HNO3 = 0.67 g / 63 g/mol = 0.0106 mol

According to stoichiometry, the ratio of HNO3 to Na2CO3 in the balanced equation is 2:1. Therefore, 0.0106 mol of HNO3 will react with an equal number of moles of Na2CO3.

Now, let's calculate the number of moles of Na2CO3 that reacts with the HNO3 in 31.0 cm3 of the HNO3 solution.

Given:
Volume of HNO3 solution used = 31.0 cm3

Using the equation:
Number of moles = concentration x volume

We know the number of moles of HNO3 (0.0106 mol) that reacted, and the volume of HNO3 solution used (31.0 cm3). The concentration of the HNO3 solution can be calculated as:

Concentration of HNO3 = Number of moles / Volume of solution
= 0.0106 mol / 31.0 cm3

Now, to find the concentration of the sodium trioxocarbonate (IV) solution, we can use the mole ratio from the balanced equation.

From the balanced equation, we can see that the ratio of Na2CO3 to HNO3 is 1:2. Therefore, the number of moles of Na2CO3 present in the reaction is half of the number of moles of HNO3, which is 0.0106 mol/2 = 0.0053 mol.

Finally, let's calculate the concentration of the sodium trioxocarbonate (IV) solution.

Given:
Volume of Na2CO3 solution = 25 cm3

Using the equation:
Concentration = Number of moles / Volume of solution

Concentration of Na2CO3 = 0.0053 mol / 25 cm3

Therefore, the concentration of the sodium trioxocarbonate (IV) solution is 0.0053 mol / 25 cm3.