25.0cm³ portions of 0.100mol in 1000cm³ solution of sodium trioxocarbonate(iv) were titrated against hydrochloric acid solution
What is the concentration of hydrochloric acid in moldm3?
The mass of concentration of sodium trioxocarbonate iv solution?
The mass of H+ formed from 1dm3 of the acid?
The mass of salt formed in the solution during titration?
Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2
1. What is the concentration of hydrochloric acid in moldm3?
Don't know and can't calculate it without knowing how much HCl was used in the titration.
2.The mass of concentration of sodium trioxocarbonate iv solution?
You had 0.1 molar Na2CO3. One mole has a mass of 1 mol which is approximately 106 grams. So you have 0.1 x 106 = ? grams Na2CO3 in the solution. But you used only 25 cc of that solution so you had moles = M x L = 0.1M x 0.025 L = moles Na2CO3. grams = moles x molar mass = ?
3. The mass of H+ formed from 1dm3 of the acid?
Don't know since we don't know the molarity of the HCl.
4. The mass of salt formed in the solution during titration?
Assuming you have enough HCl to use all of the Na2CO3 then you will have moles Na2CO3 x 2 = moles NaCl and mols NaCl x molar mass NaCl = grams NaCl.
Post your work if you get stuck. And remember to let me know the source of that funny (and incorrect) name for Na2CO3.
To find the concentration of hydrochloric acid in mol/dm³, we need to calculate the number of moles of sodium trioxocarbonate (IV) that reacted with the acid.
Given:
Volume of sodium trioxocarbonate (IV) solution = 25.0 cm³
Molarity of sodium trioxocarbonate (IV) solution = 0.100 mol/dm³
Volume of hydrochloric acid solution = 1000 cm³
Step 1: Convert the volume of sodium trioxocarbonate (IV) solution to dm³
Volume of sodium trioxocarbonate (IV) solution = 25.0 cm³ = 25.0/1000 = 0.0250 dm³
Step 2: Calculate the number of moles of sodium trioxocarbonate (IV) reacted
Moles of sodium trioxocarbonate (IV) = Molarity x Volume
Moles of sodium trioxocarbonate (IV) = 0.100 mol/dm³ x 0.0250 dm³ = 0.0025 mol
Step 3: Since the stoichiometric ratio between sodium trioxocarbonate (IV) and hydrochloric acid is 1:2, the number of moles of hydrochloric acid is twice the number of moles of sodium trioxocarbonate (IV).
Moles of hydrochloric acid = 2 x 0.0025 mol = 0.0050 mol
Step 4: Calculate the concentration of hydrochloric acid in mol/dm³
Concentration of hydrochloric acid = Moles of hydrochloric acid / Volume of hydrochloric acid solution
Concentration of hydrochloric acid = 0.0050 mol / 1.000 dm³ = 0.0050 mol/dm³
Therefore, the concentration of hydrochloric acid is 0.0050 mol/dm³.
To find the mass of sodium trioxocarbonate (IV) solution, we need to use the formula:
Mass = Volume x Density
Given:
Volume of sodium trioxocarbonate (IV) solution = 25.0 cm³
Density of sodium trioxocarbonate (IV) solution = Let's assume it is 1.00 g/cm³ (you can use the actual value if known)
Step 1: Convert the volume of sodium trioxocarbonate (IV) solution to dm³
Volume of sodium trioxocarbonate (IV) solution = 25.0 cm³ = 25.0/1000 = 0.0250 dm³
Step 2: Calculate the mass of sodium trioxocarbonate (IV) solution
Mass = Volume x Density
Mass = 0.0250 dm³ x 1.00 g/cm³ = 0.0250 kg
Therefore, the mass of sodium trioxocarbonate (IV) solution is 0.0250 kg.
To find the mass of H+ formed from 1 dm³ of the acid, we need to consider the stoichiometric ratio between hydrochloric acid and hydrogen ions. Since hydrochloric acid completely dissociates in water, it produces 1 mole of H+ ions for every mole of hydrochloric acid.
Given:
Volume of hydrochloric acid solution = 1 dm³
Step 1: Calculate the mass of H+ formed from 1 dm³ of the acid (assuming the density of hydrochloric acid is 1.00 g/cm³)
Moles of H+ = Moles of hydrochloric acid = 0.0050 mol (calculated earlier)
Step 2: Calculate the mass of H+
Mass of H+ = Moles of H+ x Molar mass of H+
The molar mass of H+ is 1 g/mol since it is just a proton.
Mass of H+ = 0.0050 mol x 1 g/mol = 0.0050 g
Therefore, the mass of H+ formed from 1 dm³ of the acid is 0.0050 g.
To find the mass of salt formed in the solution during titration, we need to know the balanced chemical equation between sodium trioxocarbonate (IV) and hydrochloric acid.
Assuming the balanced equation is:
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
From the equation, we can see that 1 mole of sodium trioxocarbonate (IV) reacts with 2 moles of hydrochloric acid to form 2 moles of sodium chloride (salt).
Given:
Moles of sodium trioxocarbonate (IV) = 0.0025 mol (calculated earlier)
Step 1: Calculate the moles of sodium chloride (salt) formed
Moles of sodium chloride = Moles of sodium trioxocarbonate (IV) x (2 moles of sodium chloride / 1 mole of sodium trioxocarbonate)
Moles of sodium chloride = 0.0025 mol x (2/1) = 0.0050 mol
Step 2: Calculate the mass of sodium chloride (salt)
Mass of sodium chloride = Moles of sodium chloride x Molar mass of sodium chloride
The molar mass of sodium chloride (NaCl) = 23.0 g/mol (atomic mass of sodium) + 35.5 g/mol (atomic mass of chlorine)
Molar mass of sodium chloride = 58.5 g/mol
Mass of sodium chloride = 0.0050 mol x 58.5 g/mol = 0.2925 g
Therefore, the mass of salt formed in the solution during titration is 0.2925 g.
To find the concentration of hydrochloric acid in mol/dm³, we need to use the concept of stoichiometry. Here are the steps to follow:
1. Determine the number of moles of sodium trioxocarbonate (Na₂CO₃) in the given volume (25.0 cm³) of the solution using the formula:
Moles of Na₂CO₃ = Volume (cm³) × Concentration (mol/dm³)
This will give you the number of moles of Na₂CO₃ in 25.0 cm³ of the solution.
2. Since sodium trioxocarbonate reacts with hydrochloric acid (HCl) in a 1:2 ratio, the number of moles of HCl will be twice the number of moles of Na₂CO₃. So, multiply the moles of Na₂CO₃ by 2 to find the number of moles of HCl.
3. Finally, divide the number of moles of HCl obtained in step 2 by the volume of the acid solution (1000 cm³) to find the concentration of HCl in mol/dm³.
To find the mass of sodium trioxocarbonate solution, we need to know the concentration (%w/v) or the density of the solution. Once we have this information, we can use the following formula:
Mass of Na₂CO₃ solution = Volume (cm³) × Density (g/cm³) or Volume (cm³) × Concentration (%w/v)
To find the mass of H⁺ formed from 1 dm³ of the acid, we need to know the molar mass of H⁺. Since H⁺ is a single proton, its molar mass is approximately 1 g/mol. Therefore, the mass of H⁺ can be calculated by multiplying the concentration of the acid (in mol/dm³) by the molar mass of H⁺ (1 g/mol).
To find the mass of salt formed in the solution during titration, we need to know the balanced chemical equation for the reaction between Na₂CO₃ and HCl. From the balanced equation, we can determine the stoichiometric ratio between Na₂CO₃ and the salt formed. With the number of moles of Na₂CO₃ determined in step 1, we can multiply it by the molar mass of the salt to find the mass of the salt formed.
Please provide the concentration (%w/v) or density of the Na₂CO₃ solution and the balanced chemical equation for a more accurate calculation of the mass of the solution and the mass of the salt formed.