If 5.30mL of 0.1000 M NaOH solution is needed to just neutralize excess acid after 20.00 mL of 0.1000 M HCl was added to 1.00 g of an antacid, how many moles of acid can the antacid counteract per gram?

20.00 x 0.1 = 2.0 millimols initially.

5.30 x 0.1 = 0.530 mmols NaOH for excess.
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2-0.53 = 1.47 millimols acid neutralized by the 1.00 g antacid tablet.
So that would be 0.00147 mols/1 g tab.

To find out how many moles of acid the antacid can counteract per gram, we need to calculate the number of moles of NaOH used to neutralize the excess acid and then divide it by the mass of the antacid.

Step 1: Calculate the number of moles of NaOH used.
Moles of NaOH = Volume of NaOH (in L) x Molarity of NaOH

Given:
Volume of NaOH = 5.30 mL = 5.30/1000 L = 0.00530 L
Molarity of NaOH = 0.1000 M

Moles of NaOH = 0.00530 L x 0.1000 mol/L

Step 2: Calculate the number of moles of HCl in excess.
Moles of HCl = Volume of HCl (in L) x Molarity of HCl

Given:
Volume of HCl = 20.00 mL = 20.00/1000 L = 0.0200 L
Molarity of HCl = 0.1000 M

Moles of HCl = 0.0200 L x 0.1000 mol/L

Step 3: Calculate the moles of HCl neutralized by NaOH.
Moles of HCl neutralized = Moles of NaOH - Moles of HCl

Step 4: Calculate the moles of acid counteracted per gram of antacid.
Moles of acid counteracted per gram = Moles of HCl neutralized / mass of antacid

Given:
Mass of antacid = 1.00 g

Moles of acid counteracted per gram = (Moles of NaOH - Moles of HCl) / 1.00 g

Now, we can substitute the values into the equation and calculate the answer.

To find out how many moles of acid the antacid can counteract per gram, we need to determine the number of moles of acid and the mass of the antacid.

First, let's calculate the number of moles of NaOH used to neutralize the HCl. The volume of NaOH used is 5.30 mL, and the molarity of NaOH is 0.1000 M.

Number of moles of NaOH = volume (in L) x molarity
= 5.30 mL x (1 L / 1000 mL) x 0.1000 M

Now, let's calculate the number of moles of HCl that was added to the antacid. The volume of HCl used is 20.00 mL, and the molarity of HCl is 0.1000 M.

Number of moles of HCl = volume (in L) x molarity
= 20.00 mL x (1 L / 1000 mL) x 0.1000 M

Since the reaction between NaOH and HCl is 1:1, the number of moles of NaOH used is equal to the number of moles of HCl added to the antacid.

Now, we need to determine the mass of the antacid. The mass given is 1.00 gram.

Finally, to calculate the moles of acid that the antacid can neutralize per gram, we divide the moles of acid by the mass of the antacid.

Moles of acid per gram = moles of acid / mass of antacid

By substituting the values, we can find the answer.

Well, it seems our antacid is quite the neutralizer! Now, let's crunch some numbers and find out how many moles of acid it can counteract per gram.

We know that 20.00 mL of 0.1000 M HCl was added, and it took 5.30 mL of 0.1000 M NaOH to neutralize the excess acid. So, the moles of HCl that reacted with NaOH can be calculated by multiplying the volume (in liters) by the molarity.

Moles of HCl = (20.00 mL / 1000 mL/L) * 0.1000 mol/L = 0.0020 mol

Since the antacid weighs 1.00 g, we can determine the moles of acid that can be counteracted per gram by simply dividing the moles of HCl by the mass in grams.

Moles of acid counteracted per gram = 0.0020 mol / 1.00 g = 0.0020 mol/g

Voila! The antacid can counteract 0.0020 moles of acid per gram. Keep in mind, though, that this calculation assumes no other reactions are taking place and all the acid is neutralized by the antacid alone.