A 0.2264 g sample of a pure carbonate, XnCO3(s) , was dissolved in 50.0 mL of 0.1800 M HCl(aq) . The excess HCl(aq) was back titrated with 24.90 mL of 0.0980 M NaOH(aq) . How many moles of HCl react with the carbonate

Answer:

6.56x10⁻³ mol

Explanation:

First we calculate how many HCl moles are there in 50.0 mL of a 0.1800 M solution:

• 0.1800 M * 50.0 mL = 9.00 mmol HCl

Then we need to calculate how many HCl moles were in excess, that number is the same as the number of NaOH moles they reacted with:

• 0.0980 M * 24.90 mL = 2.44 mmol NaOH = 2.44 mmol HCl

Finally we calculate the difference between the original number of HCl moles and the number remaining after the reaction with XnCO₃:

• 9.00 mmol - 2.44 mmol = 6.56 mmol

• 6.56 mmol / 1000 = 6.56x10⁻³ mol

Final answer:

The amount of HCl that reacted with the carbonate is found by subtracting the amount of HCl neutralized by NaOH from the total initial amount of HCl, resulting in 0.00656 moles of HCl reacting with carbonate.

Explanation:

In this problem, we first need to calculate the total amount of HCl initially used. This is done using the formulaMolarity (M) = moles of solute/Liters of solution. We can calculate moles of HCl as (0.1800 moles/L) * (0.0500 L) = 0.00900 moles.

Next, the HCl reacts with the carbonate, but there is excess that needed to be neutralized using NaOH. The amount of HCl neutralized by NaOH can be calculated similarly as (0.0980 moles/L) * (0.0249 L) = 0.00244 moles.

Thus, the amount of HCl that reacted with the carbonate is the initial moles of HCl minus the moles of HCl neutralized by NaOH. This yields 0.00900 moles - 0.00244 moles = 0.00656 moles of HCl reacted with the carbonate.

Learn more about Acid-base titration here:

brainly.com/question/34855887

#SPJ3