Question

1.4gm of mixture of CaCO3, and MgCO3 was dissolved in 200 ml of 0.2N Hcl. After the completion of reaction, the resulting solution was diluted to 250ml and 10ml of this solution required 12 ml of N/30 NaOH for neutralization calculate percentage composition of this mixture​

Answer 1

Answer: 0.893 g / 84.31

Explanation: Let's first calculate the number of moles of HCl that reacted with the mixture:

moles of HCl = 0.2 N x 0.2 L = 0.04 moles

The balanced chemical equation for the reaction between HCl and CaCO3 is:

CaCO3 + 2HCl → CaCl2 + CO2 + H2O

The balanced chemical equation for the reaction between HCl and MgCO3 is:

MgCO3 + 2HCl → MgCl2 + CO2 + H2O

We can use the number of moles of HCl and the stoichiometry of these reactions to calculate the total number of moles of CaCO3 and MgCO3 in the mixture:

moles of CaCO3 + moles of MgCO3 = 0.04 moles

Let x be the mass of CaCO3 and y be the mass of MgCO3 in the mixture. Then we can write:

x + y = 1.4 g (total mass of mixture)

x/100.09 + y/84.31 = 0.04 (total moles of mixture)

Solving these equations, we get:

x = 0.507 g (approx.)

y = 0.893 g (approx.)

Therefore, the mixture contains approximately 36.2% CaCO3 and 63.8% MgCO3 by mass.

Now, let's calculate the number of moles of NaOH required to neutralize 10 mL of the diluted solution:

moles of NaOH = (12 mL)(1/30 N)(1/1000 L/mL) = 0.0004 moles

The balanced chemical equation for the neutralization reaction between NaOH and HCl is:

NaOH + HCl → NaCl + H2O

We can use the number of moles of NaOH and the stoichiometry of this reaction to calculate the number of moles of HCl that remained in the diluted solution:

moles of HCl remaining = moles of NaOH = 0.0004 moles

The diluted solution has a volume of 250 mL, so its concentration of HCl is:

[HCl] = moles of HCl remaining / volume of solution = 0.0004 moles / 0.250 L = 0.0016 N

The 10 mL of diluted solution used for titration was taken from the original solution that was prepared by dissolving the mixture in 200 mL of 0.2 N HCl. Therefore, the concentration of HCl in the original solution is:

[HCl] = (0.2 N)(200 mL / 250 mL) = 0.16 N

Since the number of moles of HCl in the original solution is equal to the number of moles of HCl that reacted with the mixture, we can calculate the number of moles of the mixture:

moles of mixture = moles of HCl reacted = 0.04 moles

The mass of the mixture is 1.4 g, so its molar mass is:

molar mass of mixture = 1.4 g / 0.04 moles = 35 g/mol

The mass of CaCO3 in the mixture is 0.507 g, so its number of moles is:

moles of CaCO3 = 0.507 g / 100.09 g/mol = 0.005067 moles

The mass of MgCO3 in the mixture is 0.893 g, so its number of moles is:

moles of MgCO3 = 0.893 g / 84.31