Titration equations. last one for today?

A 0.425 g sample of an antacid is dissolved in water. To the mixture is added 30.00mL of 0.50 M HCl. The solution is then tritated to the endpoint with 9.82 mL of 0.50 M NaOH.

A: What volume of acid was neutralized by the NaOH?

B: What volume of acid was neutralized by the antacid?

C: What volume of acid is neutralizied per gram of antacid?

As you can already tell I am not that fluient in chemistry.
any help would be greatly appreciated.
oh and please show calculations.

Answer:
A. mL x M = mL x M
9.82 mL x 0.5 = X mL x 0.5 M
9.82 mL of HCl were neutralized by the NaOH

B. 30.00 mL less 9.82 mL neutralized by NaOH = 20.18 mL neutralized by the antacid.

C. 20.18 mL neutralized by 0.425 g, so 20.18/0.425 or 47.48 mL o 0.50 M HCl would be neutralized by 1.00 gram of antacid. .

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