CaCO3(s) + 2 HCl(aq) --> CaCl3(aq) + H2O(l) + CO2
It takes 19.26 mL of 0.6721 M HCl to completely react with the CaCO3. What is the mass of CaCO3 in the tablet?
didnt mean to cal you
First correct the formula for calcium chloride to CaCl2.
mols HCl required = M x L = ?
mols CaCO3 = 1/2 that from the coefficients.
grams CaCO3 = mols CaCO3 x molar mass CaCO3
No problem! I'm here to help. To find the mass of CaCO3 in the tablet, we can use the concept of stoichiometry and the given information.
First, let's calculate the number of moles of HCl that reacted using the given volume and concentration:
moles of HCl = volume of HCl (in liters) × concentration of HCl
To convert the volume of HCl to liters, we divide it by 1000:
volume of HCl (in liters) = 19.26 mL / 1000 = 0.01926 L
Now, let's calculate the moles of HCl:
moles of HCl = 0.01926 L × 0.6721 mol/L = 0.012952146 mol
According to the balanced equation, the reaction ratio between CaCO3 and HCl is 1:2. This means that for every 1 mole of CaCO3, 2 moles of HCl are needed.
Using this ratio, we can determine the moles of CaCO3 that reacted:
moles of CaCO3 = 1/2 × moles of HCl = 1/2 × 0.012952146 mol = 0.006476073 mol
Lastly, we need to find the molar mass of CaCO3. By adding up the atomic masses of each element in the compound (Ca: 40.08 g/mol, C: 12.01 g/mol, O: 16.00 g/mol), we get:
molar mass of CaCO3 = 40.08 g/mol + 12.01 g/mol + 3 × 16.00 g/mol = 100.09 g/mol
Now, we can calculate the mass of CaCO3 in the tablet:
mass of CaCO3 = moles of CaCO3 × molar mass of CaCO3 = 0.006476073 mol × 100.09 g/mol ≈ 0.648 g
Therefore, the mass of CaCO3 in the tablet is approximately 0.648 grams.