A TUMS tablet was dissolved in 110.00 ml of 0.1022 M HCl and analyzed for CaCO3 according to the procedure described in the lab manual. It took 7.48 ml of 0.1005 M NaOH to reach the endpoint. Calculate the mass of CaCO3 in TUMS tablet to the correct number of significant figures
CaC03(aq)+ 2HCl(aq) = CaCl2 (aq)+ H2O(l)+ CO2(g)
NaOH(aq) + HCl(aq) = NaCl(aq)+ H20(l)
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To calculate the mass of CaCO3 in the TUMS tablet, we need to use stoichiometry and the given volumes and concentrations of the HCl and NaOH solutions.
First, let's determine the moles of HCl used in the reaction. To do this, we'll use the equation provided:
2 moles of HCl react with 1 mole of CaCO3.
Given the volume and concentration of the HCl solution, we can calculate the moles of HCl:
moles of HCl = volume of HCl solution (L) × concentration of HCl solution (mol/L)
moles of HCl = 110.00 mL × 0.1022 mol/L
= 0.011252 mol
Next, we need to determine the moles of NaOH used in the reaction. From the balanced equation, we can see that it reacts in a 1:1 ratio with HCl:
1 mole of NaOH reacts with 1 mole of HCl.
Using the volume and concentration of NaOH solution, we can calculate the moles of NaOH:
moles of NaOH = volume of NaOH solution (L) × concentration of NaOH solution (mol/L)
moles of NaOH = 7.48 mL × 0.1005 mol/L
= 0.00075174 mol
Now, we can determine the moles of CaCO3 in the TUMS tablet. We know that 2 moles of HCl react with 1 mole of CaCO3:
1 mole of CaCO3 = (moles of HCl) / 2
moles of CaCO3 = 0.011252 mol / 2
= 0.005626 mol
Lastly, we need to calculate the mass of CaCO3 using its molar mass, which is:
molar mass of CaCO3 = 40.08 g/mol (Ca) + 12.01 g/mol (C) + (3 × 16.00 g/mol) (O)
molar mass of CaCO3 = 100.09 g/mol
mass of CaCO3 = moles of CaCO3 × molar mass of CaCO3
mass of CaCO3 = 0.005626 mol × 100.09 g/mol
= 0.5631 g
Therefore, the mass of CaCO3 in the TUMS tablet is 0.5631 grams.