Please Help! A particular brand of antacid contains 500 mg CaCO3 per 2.0 g table
ID: 991703 • Letter: P
Question
Please Help! A particular brand of antacid contains 500 mg CaCO3 per 2.0 g tablet according to label. 1. How many moles of CaCo3 are in one tablet? 2. The reaction by which the antacid neutralizes HCl is: 2HCl + CaCO3 > CaCl2 + CO2 + H20. How many moles of HCl can be neutralized by one tablet? 3. 50.0 ml of .300 M HCL are used to dissolve a 2.00 g tablet. How many moles of acid are used to dissolve the tablet? 4.The excess acid then requires 53.13 ml of .100 M NaOH for the back titration. How many moles of excess acid were there in the 50 ml? 5. How many moles of HCl were neutralized by the tablet?
Explanation / Answer
1) CaCO3 mass = 500 mg = 0.5 g
CaCO3 molar mass =100 g / mol
CaCO3 moles = mass / molar mass = 0.5 / 100 = 0.005
CaCO3 moles = 0.005
2)
2 HCl + CaCO3 ----------------> CaCl2 + H2O + CO2
2 mol HCl -----------------> 1 mol CaCO3
x mol HCl -----------------> 0.005 mol CaCO3
x = 0.01
moles of HCl = 0.01
3)
moles of acid = 50 x 0.300 / 1000
moles of acid = 0.015
4)
NaOH moles = 53.13 x 0.1 / 1000
NaOH moles = 5.31 x 10^-3
5)
moles of excess acid = 0.015 - 5.31 x 10^-3
moles of excess acid = 9.69 x 10^-3
6)
moles of HCl neutralised = 5.31 x 10^-3