25cm3 of 3g hydrated trioxocarbonate (iv) solution (Na2Co3. xH20) per 250cm3 was neutralised by 13.20cm3 of 0.16M Hcl solution. write the equation of reaction
h o l y
what grade is this?
You should learn to name this compound properly. It is hydrated sodium carbonate.
Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2. That's the equation. You can determine the x in the hydrate if you wish.
millimoles HCl = mL x M = 13.20 mL x 0.16 M = 2.11)rounded)
millimoles Na2CO3.xH2O = 1/2 that or 1.056 millimoles or 0.001056 moles of the compound in the 25 cc aliquot. That came from 250 cc solution so the mols in the original 250 cc is 10 x 0.001056 = 0.01056 and that was for a 3 grams sample.
mols = g/molar mass = 0.01056 = 3.00/molar mass or
molar mass = 3.00/0.01056 = 284 rounded.
Na2CO3 is 2*23 + 12 + 48 = 106 so the water is what's left or
284 - 106 = 186.4. That is 186.4/18.0 = 9.9 which rounds to 10 so the formula for the initial hydrated salt is Na2CO3.10H2O
To write the equation of the reaction between hydrated trioxocarbonate (IV) solution (Na2CO3·xH2O) and hydrochloric acid (HCl), we need to know the ratio between Na2CO3 and HCl.
First, let's calculate the moles of HCl used in the neutralization reaction:
Volume of HCl = 13.20 cm³
Concentration of HCl (Molarity) = 0.16 M
Using the formula: moles = volume × concentration
Moles of HCl = 13.20 cm³ × 0.16 mol/cm³ = 2.112 mol
Since Na2CO3 has a 1:2 ratio with HCl, we can determine the number of moles of Na2CO3 needed:
Moles of Na2CO3 = 2.112 mol / 2 = 1.056 mol
Now, let's calculate the mass of Na2CO3 used in the reaction:
Molar mass of Na2CO3 = (2 × atomic mass of Na) + atomic mass of C + (3 × atomic mass of O)
= (2 × 23.0 g/mol) + 12.0 g/mol + (3 × 16.0 g/mol)
= 46.0 g/mol + 12.0 g/mol + 48.0 g/mol
= 106.0 g/mol
Mass of Na2CO3 = moles × molar mass = 1.056 mol × 106.0 g/mol = 111.936 g
Now we can write the balanced equation for the reaction:
2 Na2CO3 + HCl → 2 NaCl + H2O + CO2
This equation shows that 2 moles of Na2CO3 react with 1 mole of HCl to produce 2 moles of NaCl, 1 mole of H2O, and 1 mole of CO2.
To write the equation of the reaction, we need to understand the components involved and their reactants. In this case, we have a 3g hydrated trioxocarbonate (IV) solution (Na2CO3.xH2O) being neutralized by a 0.16M HCl solution.
First, let's determine the molecular formula of the hydrated trioxocarbonate (IV) solution (Na2CO3.xH2O). We are given that 3g of this solution is present in 25cm3.
To find the number of moles of Na2CO3 in the solution, we need to convert grams to moles using the molar mass of Na2CO3. The molar mass of Na2CO3 is calculated as follows:
Na (sodium) has a molar mass of 22.99 g/mol
C (carbon) has a molar mass of 12.01 g/mol
O (oxygen) has a molar mass of 16.00 g/mol
Na2CO3 has two sodium atoms (22.99 g/mol * 2 = 45.98 g/mol), one carbon atom (12.01 g/mol), and three oxygen atoms (16.00 g/mol * 3 = 48.00 g/mol).
Adding these masses together, we get the molar mass of Na2CO3: 45.98 g/mol + 12.01 g/mol + 48.00 g/mol = 105.99 g/mol.
Now, we can calculate the number of moles of Na2CO3 by dividing the mass of Na2CO3 by its molar mass:
Number of moles = mass / molar mass
Number of moles = 3g / 105.99g/mol ≈ 0.02825 mol
Since we have 25cm3 of the 3g hydrated trioxocarbonate (IV) solution, and the density of water is 1 g/cm3, we can assume that the volume of the 3g hydrated trioxocarbonate (IV) solution is the same as the volume of water in the solution. Therefore, the volume of water (in cm3) in the solution is:
Volume of water = Total volume of solution - Volume of Na2CO3
Volume of water = 25cm3 - 25cm3 = 0cm3
From the given information, 13.20cm3 of 0.16M HCl solution neutralizes the solution. This implies that the number of moles of HCl used is given by:
Number of moles of HCl = concentration (M) x Volume (L)
Number of moles of HCl = 0.16mol/L x (13.20cm3 /1000cm3/L) = 0.002112 mol
Now that we have the number of moles of Na2CO3 and HCl, we can write the balanced equation for the reaction.
The balanced equation is as follows:
2Na2CO3 + H2SO4 → Na2SO4 + CO2 + H2O
However, since the reaction is with HCl, we need to modify the equation to fit our reactants. HCl is a strong acid that completely dissociates in water, so it can be represented as H+ and Cl- ions. Therefore, the equation should be:
2Na2CO3 + 2HCl → 2NaCl + CO2 + H2O
This equation represents the reaction between 2 moles of Na2CO3 and 2 moles of HCl, producing 2 moles of NaCl, 1 mole of CO2, and 1 mole of H2O.
Please note that the equation may not be balanced in terms of the number of hydrogen, oxygen, and sodium atoms, but it reflects the stoichiometry of the reaction in terms of moles.