A solution of trioxonitrate(V) acid contained 0.67g in 100cm^3. 31.0cm^3 of this solution neutralized 2.5cm^3 of a sodium trioxocarbonate(IV) solution. calculate the concentration of the trioxocarbonate(IV) solution.
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To solve this problem, we can use the concept of molarity and stoichiometry. Here are the step-by-step calculations:
Step 1: Calculate the number of moles of trioxonitrate(V) acid:
Mass = 0.67g
Volume = 100cm³
First, convert the mass to moles using the molar mass of trioxonitrate(V) acid:
Molar mass of HNO₃ = 1(1.01) + 1(14.01) + 3(16) = 63.01g/mol
Number of moles = Mass / Molar mass
Number of moles = 0.67g / 63.01g/mol = 0.0106 mol
Step 2: Calculate the concentration of trioxonitrate(V) acid solution:
Concentration = Number of moles / Volume
Concentration = 0.0106 mol / 100 cm³ = 0.0106 mol/dm³ = 0.0106 M
Step 3: Determine the stoichiometry between trioxonitrate(V) acid and sodium trioxocarbonate(IV):
From the equation:
2HNO₃ + Na₂CO₃ → 2NaNO₃ + CO₂ + H₂O
The stoichiometry is 2:1 between HNO₃ and Na₂CO₃. This means that 2 moles of HNO₃ react with 1 mole of Na₂CO₃.
Step 4: Calculate the number of moles of sodium trioxocarbonate(IV) that reacted:
Volume of HNO₃ solution = 31.0 cm³
Volume of Na₂CO₃ solution = 2.5 cm³
Using the stoichiometry, we can write:
2 mol HNO₃ → 1 mol Na₂CO₃
0.0106 mol HNO₃ → x mol Na₂CO₃
x = 0.0106 mol HNO₃ * (1 mol Na₂CO₃ / 2 mol HNO₃) = 0.0053 mol Na₂CO₃
Step 5: Calculate the concentration of the sodium trioxocarbonate(IV) solution:
Concentration = Number of moles / Volume
Concentration = 0.0053 mol / 2.5 cm³ = 0.00212 mol/dm³ = 0.00212 M
Therefore, the concentration of the sodium trioxocarbonate(IV) solution is 0.00212 M.
To calculate the concentration of the sodium trioxocarbonate(IV) solution, we can use the concept of stoichiometry from the neutralization reaction.
First, let's write out the balanced chemical equation for the reaction:
2HNO₃ + Na₂CO₃ → 2NaNO₃ + H₂O + CO₂
From the equation, we can see that 2 moles of HNO₃ react with 1 mole of Na₂CO₃. This means that the mole ratio of HNO₃ to Na₂CO₃ is 2:1.
Now, let's calculate the number of moles of trioxonitrate(V) acid using the given mass and molar mass:
Mass of trioxonitrate(V) acid = 0.67g
Molar mass of HNO₃ = 63.01g/mol
Number of moles of HNO₃ = Mass / Molar mass
= 0.67g / 63.01g/mol
≈ 0.0106 mol
Next, we'll determine the number of moles of Na₂CO₃ that reacted with the HNO₃. Since the mole ratio of HNO₃ to Na₂CO₃ is 2:1, the number of moles of Na₂CO₃ will be half the number of moles of HNO₃.
Number of moles of Na₂CO₃ = 0.0106 mol / 2
= 0.0053 mol
Now, let's calculate the concentration of the sodium trioxocarbonate(IV) solution using the volume and the number of moles of Na₂CO₃:
Volume of sodium trioxocarbonate(IV) solution = 2.5cm³
Number of moles of Na₂CO₃ = 0.0053 mol
Concentration (molarity) = Number of moles / Volume
Concentration of the sodium trioxocarbonate(IV) solution = 0.0053 mol / 2.5cm³
= 0.00212 mol/cm³ or 2.12 mol/dm³
Therefore, the concentration of the sodium trioxocarbonate(IV) solution is 2.12 mol/dm³.