What volume of 10 percentage (mass/volume) solution of Na2CO3 will be required to neutralise 100mL of HCl sution containing 3.65g of HCl.

To determine the volume of the 10% Na2CO3 solution required to neutralize the given amount of HCl, we need to use balanced chemical equations and stoichiometry.

First, write the balanced chemical equation for the reaction between Na2CO3 (sodium carbonate) and HCl (hydrochloric acid):

2NaHCO3 + HCl → Na2CO3 + 2H2O + CO2

From the balanced equation, we can see that 1 mole of Na2CO3 reacts with 2 moles of HCl.

Next, calculate the number of moles of HCl in the given solution using its mass:

Mass of HCl = 3.65 g

Now, we need to convert the mass of HCl to moles using its molar mass:

Molar mass of HCl = 36.46 g/mol

Moles of HCl = Mass of HCl / Molar mass of HCl
= 3.65 g / 36.46 g/mol
≈ 0.1001 mol

Since 1 mole of Na2CO3 reacts with 2 moles of HCl, we need twice the number of moles of HCl for Na2CO3:

Moles of Na2CO3 = 0.1001 mol × 2
= 0.2002 mol

Finally, we can calculate the volume of the 10% Na2CO3 solution required using its concentration:

Concentration of Na2CO3 solution = 10% = 10 g/100 mL

Now, we need to convert grams to moles using the molar mass of Na2CO3:

Molar mass of Na2CO3 = 105.99 g/mol

Moles of Na2CO3 = Mass of Na2CO3 / Molar mass of Na2CO3
= (10 g / 100) / 105.99 g/mol
= 0.009417 mol

To find the volume of the 10% Na2CO3 solution, we can use the equation:

Volume of Na2CO3 solution (in mL) = Moles of Na2CO3 / Concentration of Na2CO3 solution

Volume of Na2CO3 solution = (0.2002 mol / 0.009417 mol) × 100 mL
≈ 2127 mL

Therefore, approximately 2127 mL of the 10% Na2CO3 solution will be required to neutralize 100 mL of the HCl solution containing 3.65 g of HCl.

2HCl + Na2CO3 == 2NaCl + H2O + CO2

You have 3.65 g HCl. How many mols is that? That's mols = g/molar mass = 3.65/36.5 = 0.1 Look at the coefficients in the balanced equation. 0.1 mol HCl will need 1/2 that or 0.05 mols Na2CO3. How much is that in grams. mols Na2CO3 = grams Na2CO3/molar mass Na2CO3. You know molar mass and you know mols, solve for grams. Check my work.