# Chemistry

posted by Caroline

Instead of using ratios for back titrations we can also use molarities if our solutions are standardized. A 0.188g sample of antacid containing an unknown amount of triprotic base Al(OH)3 was reacted with 25.0mL of 0.101M HCl. The resulting solution was then titrated with 10.55mL of 0.132M NaOH solution. Calculate the mass percent of Al(OH)3 in the antacid sample.

This is what I have done so far:
0.025L | 0.101mol/L = 0.00252 mol HCl 0.01055 L | 0.132mol/L = 0.001393 mol NaOH 0.00252-0.001393 = 0.001127 mol HCl neutralized by antacid 3:1 ratio 0.000376 mol Al(OH)3

I am confused as to where I go from finding mols of Al(OH)3 and don't understand how to find the total to take a percent from? This is probably really dumb but I can't figure it out.

## Similar Questions

1. ### college

1) You are given solutions of HCl and NaOH and must determine their concentrations. You use 37.0mL of NaOH to titrate 100mL of HCl and 13.6 mL of NaOH to titrate 50.0mL of 0.0782 M H2SO4. Find the unknown concentrations. Molarity of …
2. ### Chemistry

1) You are given solutions of HCl and NaOH and must determine their concentrations. You use 37.0mL of NaOH to titrate 100mL of HCl and 13.6 mL of NaOH to titrate 50.0mL of 0.0782 M H2SO4. Find the unknown concentrations. Molarity of …
3. ### Chemistry

Sorry DrBob, it's me again! Here's the info: Antacid Brand: Life Concentration of HCl: 0.1845 M Concentration of NaOH: 0.1482 M Trial 1: 1.Mass of Table: 1.2173g 2.Volume of HCl added: 75.0 mL 3.Milliomoles of HCl added:(0.1845M x …
4. ### Chemistry

Sorry DrBob, it's me again! Here's the info: Antacid Brand: Life Concentration of HCl: 0.1845 M Concentration of NaOH: 0.1482 M Trial 1: 1.Mass of Table: 1.2173g 2.Volume of HCl added: 75.0 mL 3.Milliomoles of HCl added:(0.1845M x …
5. ### chemistry

31.0mL sample of LiOH, concentration unknown, is titrated with .52 M HCl. It takes 42.9 ml of HCl to reach the endpoint. What is the concentration of the base?
6. ### Chemistry

4. A 5.309 g antacid tablet, with CaCO3 as the active ingredient, was mixed was 30.0 mL of 0.831 M HCl. It took 15.3 mL of 0.7034 M NaOH to neutralize the excess acid. a) Calculate the volume of HCl neutralized by the NaOH. b) Calculate …
7. ### chemistry

0.688 g of antacid was treated with 50.00 mL of 0.100 M HCl. The excess acid required 4.21 mL of 1.200 M NaOH for back titration. What is the neutralizing power of this antacid expressed as mmol(millimoles) of HCl per gram of antacid?
8. ### Chemistry

0.688g of antacid was treated with 50.00ml of 0.200M HCl. The excess acid required 4.21 mL of 1.200 M NaOH for back-titration. What is the neutralizing power of this antacid expressed as mmol of HCl per gram of antacid?
9. ### Chemistry

Instead of using ratios for back titrations we can also use molarities if our solutions are standardized. A 0.188g sample of antacid containing an unknown amount of triprotic base Al(OH)3 was reacted with 25.0mL of 0.101M HCl. The …
10. ### Chemistry

How would the following affect the calculated value of neutralizing power of antacid (make it higher, lower or no effect)?

More Similar Questions