1.06g of sodium trioxocarbonate(iv) reacted with excess dilute 0.1M hydrochloric acid, after the reaction, the unreacted acid required 24cm3 of 0.1M sodium hydroxide for its complete neutralization. Calculate the original volume of the hydrochloric acid used.
First you need to know that Na2CO3 is sodium carbonate and not that funny name you have. The reaction with HCl is
Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2
mols Na2CO3 = grams/molar mass = 1.06/106 = 0.01 mols.]
It will take twice that number of mols to react EXACTLY with the HCl so that is 0.02 mols HCl needed for the 1.06 g Na2CO3.
The excess HCl required 24 cc of 0.1 M NaOH or mols NaOH = 0.024 x 0.1 = 0.0024 mols.
NaOH + HCl ==> NaCl + H2O
0.0024 mols NaOH must take 0.0024 mols HCl for the excess SO you must have had the 0.02 mols initially plus the 0.0024 excess or a total to start with of 0.0024 + 0.02 = 0.0224 mols. Then M = mols/L. You know mols and M, solve for L and conertr to mL. Post your work if you get stuck.
Correct, would like to learn more.
To determine the original volume of the hydrochloric acid used, we need to consider the balanced chemical equation for the reaction between sodium trioxocarbonate(IV) (Na2CO3) and hydrochloric acid (HCl):
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
From the balanced equation, we can see that 1 mole of Na2CO3 reacts with 2 moles of HCl. Therefore, we need to calculate the number of moles of Na2CO3 that reacted.
Given that the mass of Na2CO3 is 1.06g, we can calculate the number of moles of Na2CO3 using its molar mass:
Molar mass of Na2CO3 = 2(23.0) + 12.0 + 3(16.0) = 106.0 g/mol
Number of moles of Na2CO3 = mass / molar mass = 1.06g / 106.0 g/mol = 0.01 mol
Since Na2CO3 and HCl react in a 1:2 ratio, we have twice the number of moles of HCl compared to Na2CO3. Therefore, the number of moles of HCl is 2 times the number of moles of Na2CO3.
Number of moles of HCl = 2 * 0.01 mol = 0.02 mol
Now, we can use the given information about the unreacted acid requiring 24cm3 of 0.1M sodium hydroxide for complete neutralization.
From the balanced equation, we can see that 2 moles of HCl react with 2 moles of NaOH. Therefore, the number of moles of HCl is equal to the number of moles of NaOH used to neutralize the unreacted HCl.
Number of moles of NaOH = concentration * volume
Number of moles of NaOH = 0.1 mol/L * 0.024 L = 0.0024 mol
Since the number of moles of HCl is equal to the number of moles of NaOH, we have:
Number of moles of HCl = 0.0024 mol
Now, we can calculate the original volume of the hydrochloric acid used:
Original volume of HCl = number of moles / concentration
Original volume of HCl = 0.0024 mol / 0.1 mol/L = 0.024 L = 24 cm3
Therefore, the original volume of the hydrochloric acid used is 24 cm3.
To solve this problem, we can use the concept of stoichiometry to calculate the original volume of hydrochloric acid used.
Step 1: Balance the chemical equation for the reaction between sodium trioxocarbonate(IV) (also known as sodium carbonate) and hydrochloric acid:
Na2CO3 + 2HCl -> 2NaCl + H2O + CO2
Step 2: Calculate the moles of sodium trioxocarbonate(IV) used:
Given mass of sodium trioxocarbonate(IV) = 1.06g
Molar mass of sodium trioxocarbonate(IV) (Na2CO3) = 2*23 + 12 + 3*16 = 106g/mol
Moles of sodium trioxocarbonate(IV) = mass / molar mass
= 1.06g / 106g/mol
= 0.01 mol
Step 3: Determine the stoichiometric ratio between sodium trioxocarbonate(IV) and hydrochloric acid:
From the balanced equation, we see that 1 mol of sodium trioxocarbonate(IV) reacts with 2 moles of hydrochloric acid.
Step 4: Calculate the moles of hydrochloric acid reacted:
Moles of hydrochloric acid = 0.01 mol * 2
= 0.02 mol
Step 5: Calculate the volume of 0.1 M hydrochloric acid used:
Volume of hydrochloric acid (in liters) = moles of hydrochloric acid / concentration of hydrochloric acid
= 0.02 mol / 0.1 mol/L
= 0.2 L
Step 6: Convert the volume to cm3:
Volume of hydrochloric acid (in cm3) = Volume of hydrochloric acid (in liters) * 1000
= 0.2 L * 1000
= 200 cm3
Therefore, the original volume of hydrochloric acid used is 200 cm3.