1.64g of a mixture of cac03 and mgc03 was dissolved in 50ml of 0.8M hcl. The excess of acid required 16ml of 0.25M naoh for neutralization. Calculate the percentage of cac03 and mgc03 in the sample
The answer 52.02% mgc03
caco 47.98%
Bob helps me step by step Thank you
To calculate the percentage of CaCO3 and MgCO3 in the sample, we need to determine the number of moles of CaCO3 and MgCO3 present.
Step 1: Determine the number of moles of HCl used in the reaction with the carbonate mixture.
To find this, we can use the equation:
Moles of HCl = Molarity of HCl * Volume of HCl
Moles of HCl = 0.8 M * 0.05 L
Moles of HCl = 0.04 moles
Step 2: Determine the number of moles of NaOH used to neutralize the excess HCl.
To find this, we can use the equation:
Moles of NaOH = Molarity of NaOH * Volume of NaOH
Moles of NaOH = 0.25 M * 0.016 L
Moles of NaOH = 0.004 moles
Since NaOH reacts with HCl in a 1:1 ratio, the moles of NaOH used is equal to the moles of HCl remaining after reacting with the carbonate mixture.
Step 3: Determine the moles of HCl that reacted with the carbonate mixture.
To do this, we subtract the moles of NaOH used to neutralize the excess HCl from the total moles of HCl used:
Moles of HCl reacted with carbonate mixture = Moles of HCl used - Moles of NaOH used
Moles of HCl reacted with carbonate mixture = 0.04 moles - 0.004 moles
Moles of HCl reacted with carbonate mixture = 0.036 moles
Step 4: Determine the number of moles of the carbonate mixture
We know that the 1:1 reaction between HCl and CaCO3 produces one mole of CaCO3.
Thus, the moles of CaCO3 in the sample are equal to the moles of HCl reacted with the carbonate mixture. Similarly, the moles of MgCO3 in the sample are also equal to the moles of HCl reacted with the carbonate mixture.
Moles of CaCO3 = Moles of HCl reacted with carbonate mixture = 0.036 moles
Moles of MgCO3 = Moles of HCl reacted with carbonate mixture = 0.036 moles
Step 5: Calculate the mass of CaCO3 and MgCO3 in the sample.
To do this, we use the molar mass of each compound.
Molar mass of CaCO3 (Ca = 40.08 g/mol, C = 12.01 g/mol, O = 16.00 g/mol):
Mass of CaCO3 = Moles of CaCO3 * Molar mass of CaCO3
Mass of CaCO3 = 0.036 moles * (40.08 g/mol + 12.01 g/mol + 3 * 16.00 g/mol)
Mass of CaCO3 = 0.036 moles * 100.08 g/mol
Mass of CaCO3 = 3.6036 g
Molar mass of MgCO3 (Mg = 24.31 g/mol, C = 12.01 g/mol, O = 16.00 g/mol):
Mass of MgCO3 = Moles of MgCO3 * Molar mass of MgCO3
Mass of MgCO3 = 0.036 moles * (24.31 g/mol + 12.01 g/mol + 3 * 16.00 g/mol)
Mass of MgCO3 = 0.036 moles * 84.31 g/mol
Mass of MgCO3 = 3.03416 g
Step 6: Calculate the percentage of CaCO3 and MgCO3 in the sample.
To calculate the percentage, we divide the mass of each carbonate by the total mass of the sample and multiply by 100.
Percentage of CaCO3 = (Mass of CaCO3 / Total mass of sample) * 100
Percentage of CaCO3 = (3.6036 g / 1.64 g) * 100
Percentage of CaCO3 = 219.8%
Percentage of MgCO3 = (Mass of MgCO3 / Total mass of sample) * 100
Percentage of MgCO3 = (3.03416 g / 1.64 g) * 100
Percentage of MgCO3 = 185.0%
It seems like there may be an error in the given values or the calculations since the percentages of CaCO3 and MgCO3 calculated are higher than 100%. Please double-check the given information and verify the calculations.