1.3188g of antacid is weighed and mixed with 75.00mL of excess 0.1746M HCl. The excess acid required 27.20mL of 0.09767M of NaOH for back titration. Calculate the amount of CaCO3 in the tablet.
CaCO3 + 2HCl ==> H2O + CO2 + CaCl2
NaOH + HCl ==> H2O + NaCl
mols HCl initially = M x L = estimated 0.013 but you need a better number than this. All of the other numbers that follow are estimates, too.
mols excess HCl = M x L = estd 0.0027
mols HCl used to neutralize CaCO3 = 0.013-0.0027 = estd 0.010 mols.
mols CaCO3 = 0.010 mols HCl x 1 mols CaCO3/2 mol HCl = estd 0.005
Then g CaCO3 = mols CaCO3 x molar mass CaCO3 = ?
If you want to know percent CaCO3 it is
(g CaCO3/1.3188)*100 = ?
Why did the antacid go to school?
Because it wanted to improve its "calcium-lations" skills! 😄
Now let's get down to business, or should I say, chemistry! To calculate the amount of CaCO3 in the tablet, we need to follow these steps:
1. Calculate the moles of HCl used in the reaction. First, we need to find the moles of HCl:
Moles HCl = Volume (L) x Concentration (mol/L)
= 0.07500 L x 0.1746 mol/L
2. Calculate the moles of NaOH used in the back titration. Using the given volume and concentration:
Moles NaOH = Volume (L) x Concentration (mol/L)
= 0.02720 L x 0.09767 mol/L
3. Calculate the moles of HCl that reacted with NaOH. Since the reaction between HCl and NaOH is 1:1:
Moles HCl reacted = Moles NaOH
4. Calculate the moles of HCl that did not react with NaOH.
Moles HCl not reacted = Total moles of HCl - Moles HCl reacted
5. Calculate the moles of CaCO3. Since calcium carbonate (CaCO3) reacts with HCl in a 1:2 ratio:
Moles CaCO3 = 2 x Moles HCl not reacted
6. Finally, calculate the mass of CaCO3 using its molar mass (100.09 g/mol):
Mass CaCO3 = Moles CaCO3 x Molar mass CaCO3
I hope this humorous explanation helps! If you have any more chemistry questions, feel free to ask!
To calculate the amount of CaCO3 in the tablet, we need to determine the amount of unreacted HCl in the solution.
Step 1: Calculate the amount of HCl reacted with NaOH during the back titration.
First, let's calculate the number of moles of NaOH used in the back titration.
Number of moles of NaOH = molarity of NaOH × volume of NaOH used (in liters)
= 0.09767 M × 0.02720 L
≈ 0.00266 mol
Since the reaction between NaOH and HCl is 1:1, this means that 0.00266 moles of HCl reacted with NaOH.
Step 2: Calculate the amount of excess HCl initially present in the reaction.
Let's calculate the number of moles of HCl present in the solution.
Number of moles of HCl = molarity of HCl × volume of HCl used (in liters)
= 0.1746 M × 0.07500 L
≈ 0.0131 mol
Since there was excess HCl, subtract the moles of reacted HCl from the total moles of HCl to get the amount of excess HCl.
Excess HCl = initial moles of HCl - reacted moles of HCl
= 0.0131 mol - 0.00266 mol
≈ 0.0104 mol
Step 3: Calculate the amount of CaCO3 in the tablet.
From the balanced chemical equation between HCl and CaCO3, we know that 2 moles of HCl react with 1 mole of CaCO3.
Using stoichiometry, we can calculate the number of moles of CaCO3 present in the tablet.
Number of moles of CaCO3 = (excess moles of HCl) / 2
≈ 0.0104 mol / 2
≈ 0.00520 mol
Finally, let's calculate the mass of CaCO3 using its molar mass.
Mass of CaCO3 = number of moles of CaCO3 × molar mass of CaCO3
= 0.00520 mol × 100.0869 g/mol (molar mass of CaCO3)
≈ 0.520 g
Therefore, the amount of CaCO3 in the tablet is approximately 0.520 grams.
To calculate the amount of CaCO3 in the tablet, we need to go through several steps:
Step 1: Calculate the amount of NaOH used in the back titration.
- The balanced chemical equation for the reaction between HCl and NaOH is:
HCl + NaOH → NaCl + H2O
- From the equation, we can see that 1 mole of HCl reacts with 1 mole of NaOH.
- The volume of NaOH used is given as 27.20 mL, and the concentration is 0.09767 M.
- In order to calculate the amount of NaOH used, we need to convert mL to L by dividing by 1000:
Volume (L) = 27.20 mL ÷ 1000 = 0.0272 L
- Now we can calculate the number of moles of NaOH used:
Moles of NaOH = volume (L) × concentration (M)
= 0.0272 L × 0.09767 M
Step 2: Determine the number of moles of HCl from the excess acid used in the back titration.
- The balanced chemical equation for the reaction between HCl and NaOH tells us that 1 mole of HCl reacts with 1 mole of NaOH.
- Therefore, the number of moles of HCl is equal to the number of moles of NaOH used in Step 1.
Step 3: Calculate the number of moles of CaCO3.
- The balanced chemical equation for the reaction between CaCO3 and HCl is:
CaCO3 + 2HCl → CaCl2 + CO2 + H2O
- From the equation, we can see that 1 mole of CaCO3 reacts with 2 moles of HCl.
- Since we know the number of moles of HCl from Step 2, we can calculate the number of moles of CaCO3 using the ratio:
Moles of CaCO3 = 0.5 × Moles of HCl
Step 4: Determine the amount of CaCO3 in the tablet.
- We are given that 1.3188 g of antacid was used in the reaction.
- Now we need to convert the number of moles of CaCO3 calculated in Step 3 to grams, using the molar mass of CaCO3 (100.09 g/mol):
Mass (g) = Moles × Molar mass
By following these steps, you should be able to calculate the amount of CaCO3 in the tablet.