Question

When a student analyzed a different tablet of an antacid massing 1.4871 g. 25.00 mL of 0.7004M HCl were used to dissolve the tablet. The HCl remaining after the reaction required 6.85 mL of 1.108 M NaOH to back titrate. How many moles of HCI did the tablet neutralize?

Select one:
A. 0.007590 mol
B. 0.01751 mol
C. 0.7004 mol
D. 0.009920 mol

Answer 1

The number of moles of HCl that the antacid tablet neutralized is 0.009920 mol. This is found by calculating the total moles of HCl added and subtracting the moles of HCl that were neutralized by the NaOH in the back titration.

Step-by-step explanation:

To calculate the number of moles of HCl that the antacid tablet neutralized, first, we need to determine the total moles of HCl initially added. Then we subtract the moles of HCl that were neutralized by the NaOH during the back titration.

The total moles of HCl initially added is calculated as follows:

Total moles of HCl = Volume of HCl × Molarity of HCl

Total moles of HCl = 25.00 mL × 0.7004 M

Converting mL to L (since 1 mL = 0.001 L), we have:

Total moles of HCl = 0.02500 L × 0.7004 M = 0.01751 mol of HCl

Next, we calculate the moles of HCl that reacted with NaOH:

Moles of HCl reacted with NaOH = Volume of NaOH × Molarity of NaOH

Moles of HCl reacted with NaOH = 6.85 mL × 1.108 M

Converting mL to L gives:

Moles of HCl reacted with NaOH = 0.00685 L × 1.108 M = 0.007590 mol of HCl

Finally, to find how many moles of HCl the antacid neutralized, we subtract the moles of HCl that reacted with NaOH from the total moles of HCl:

Moles of HCl neutralized by the tablet = Total moles of HCl - Moles of HCl reacted with NaOH

Moles of HCl neutralized by the tablet = 0.01751 mol - 0.007590 mol = 0.009920 mol

So, the correct answer is D. 0.009920 mol.