If 5.05mL of 0.1000 M NaOH solution is needed to just neutralize excess acid after 20.00 mL of HCl was added to 1.00 g of an antacid, how many moles of acid can the antacid counteract per gram?
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You need to know the M of the HCl used.
millimoles HCl used = 20.00 x ?M = xx.
Excess is 5.05 mL x 0.1000 M NaOH = 0.505 mmoles excess acid.
Initial mmoles - excess mmoles = mmoles antacid in the 1.00 g tablet. Then convert mmoles to moles.
To find the number of moles of acid that the antacid can counteract per gram, we need to determine the number of moles of NaOH that were required to neutralize the excess acid.
Step 1: Calculate the number of moles of NaOH used.
Using the equation: moles = concentration x volume, we can calculate the number of moles of NaOH used.
moles of NaOH = 0.1000 M x 5.05 mL = 0.00505 moles
Step 2: Convert the number of moles of NaOH used to moles of HCl.
Since the reaction between NaOH and HCl is 1:1, the number of moles of HCl used is also 0.00505 moles.
Step 3: Calculate the number of grams of HCl used.
Using the equation: mass = moles x molar mass, we can calculate the number of grams of HCl used.
molar mass of HCl = 1 g/mol + 35.5 g/mol = 36.5 g/mol
mass of HCl = 0.00505 moles x 36.5 g/mol = 0.184 g
Step 4: Calculate the number of grams of antacid.
Given that the mass of the antacid is 1.00 g, we have the mass of the antacid.
Step 5: Calculate the number of moles of acid that the antacid can counteract per gram.
Using the equation: moles of acid per gram = moles of HCl / mass of antacid, we can calculate the number of moles of acid that the antacid can counteract per gram.
moles of acid per gram = 0.184 moles / 1.00 g = 0.184 mol/g
Therefore, the antacid can counteract 0.184 moles of acid per gram.
To find the number of moles of acid that the antacid can counteract per gram, we need to determine the number of moles of acid neutralized by the NaOH solution.
First, let's calculate the number of moles of NaOH used in the neutralization:
Moles of NaOH = concentration of NaOH solution (in M) × volume of NaOH used (in L)
Given:
Concentration of NaOH solution = 0.1000 M (moles per liter)
Volume of NaOH used = 5.05 mL = 5.05/1000 L (converted to liters)
Substituting the values into the formula:
Moles of NaOH = 0.1000 M × 5.05/1000 L = 0.00505 moles
Since the stoichiometry of the reaction is 1:1 between NaOH and HCl, the moles of NaOH used is equal to the moles of HCl neutralized. Therefore, the number of moles of HCl neutralized is also 0.00505 moles.
Now, let's calculate the number of moles of HCl per gram of antacid:
Moles of HCl per gram of antacid = Moles of HCl / Mass of antacid (in grams)
Given:
Mass of antacid = 1.00 g
Substituting the values into the formula:
Moles of HCl per gram of antacid = 0.00505 moles / 1.00 g = 0.00505 moles/g
Therefore, the antacid can neutralize 0.00505 moles of acid per gram.