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 mgco3.
The right answer mgc03 52.02%
caco 47.98%
Who help for me step by step
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To solve this problem, we need to use the concept of stoichiometry and the law of conservation of mass. Here's how you can approach the problem step by step:
1. Calculate the number of moles of HCl used:
- We are given the volume and concentration of HCl. Use the formula C = n/V, where n is the number of moles and V is the volume.
- Convert the volume to liters: 50 ml = 0.05 L.
- Calculate the number of moles using the formula n = C * V.
2. Calculate the number of moles of NaOH used:
- Again, we are given the volume and concentration of NaOH.
- Convert the volume to liters: 16 ml = 0.016 L.
- Calculate the number of moles using the formula n = C * V.
3. Determine the limiting reactant:
- To find out which compound (CaCO3 or MgCO3) is in excess, we need to compare the moles of HCl and NaOH used.
- Divide the number of moles of each compound by their respective stoichiometric coefficients in the neutralization reaction:
- 1 mole of HCl reacts with 1 mole of CaCO3.
- 2 moles of HCl react with 1 mole of MgCO3 (according to the balanced equation).
4. Calculate the moles of CaCO3 and MgCO3:
- Multiply the moles of HCl used by the stoichiometric coefficients for each compound.
5. Calculate the mass of CaCO3 and MgCO3:
- Multiply the number of moles of each compound by their respective molar masses:
- Molar mass of CaCO3 = 40.08 g/mol (Ca = 40.08 g/mol, C = 12.01 g/mol, O = 16.00 g/mol).
- Molar mass of MgCO3 = 84.32 g/mol (Mg = 24.31 g/mol, C = 12.01 g/mol, O = 16.00 g/mol).
6. Calculate the percentage of CaCO3 and MgCO3:
- Divide the mass of each compound by the total mass of the mixture (given as 1.64 g) and multiply by 100 to get the percentage.
Following these steps, you should be able to calculate the percentage of CaCO3 and MgCO3 in the mixture.