How do I calculate the mass of sodium carbonate required to neutralize 15-20 mL
of 0.25M HCl at the stoichiometric point?
Na2CO2 + 2HCl ==> 2NaCl + CO2 + H2O
Say 20 mL. How many mols is that? M x L = mols HCl.
mols Na2CO3 = 1/2 that.
g Na2CO3 = mols x molar mass.
To calculate the mass of sodium carbonate required to neutralize 15-20 mL of 0.25M HCl at the stoichiometric point, you need to understand the concept of stoichiometry and the balanced chemical equation for the reaction between sodium carbonate (Na2CO3) and hydrochloric acid (HCl).
The balanced chemical equation for the reaction is:
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
From the balanced equation, we can see that 1 mole of sodium carbonate (Na2CO3) reacts with 2 moles of hydrochloric acid (HCl). This means that the ratio of Na2CO3 to HCl is 1:2.
To find the number of moles of HCl present in the given volume of 15-20 mL (assuming it is the same range for calculation purposes), we can use the formula:
moles of HCl = concentration (M) × volume (L)
Given that the concentration of HCl is 0.25M and the volume is 15-20 mL (which is 0.015-0.020 L), we can calculate the range of moles of HCl present.
moles of HCl = 0.25M × (0.015-0.020 L) = 0.00375-0.005 moles
Since the ratio of Na2CO3 to HCl is 1:2, the number of moles of Na2CO3 needed to react with the range of moles of HCl is the same.
moles of Na2CO3 = 0.00375-0.005 moles
To calculate the mass of sodium carbonate required, we need to use the molar mass of Na2CO3. The molar mass of sodium carbonate is:
Na (22.99 g/mol) + C (12.01 g/mol) + 3O (16.00 g/mol) = 105.99 g/mol
Now we can calculate the mass of sodium carbonate using the range of moles calculated:
mass of Na2CO3 = moles of Na2CO3 × molar mass of Na2CO3
mass of Na2CO3 = (0.00375-0.005 moles) × 105.99 g/mol
After performing the calculation, you will get the mass of sodium carbonate required to neutralize 15-20 mL of 0.25M HCl at the stoichiometric point.