A mixture consists of lithium carbonate (Li2CO3) and potassium carbonate (K2CO3)

Question

A mixture consists of lithium carbonate (Li2CO3) and potassium carbonate (K2CO3). These react with hydrochloric acid as follows.
Li2CO3(s) + 2HCl(aq)2LiCl(aq) + H2O + CO2(g)
K2CO3(s) + 2HCl(aq)2KCl(aq) + H2O + CO2(g)
When8.982g of this mixture was analyzed, it consumed95.12mL of2.043M HCl.

1)Calculate the number of grams of lithium carbonate in the original mixture
2)Calculate the number of grams of potassium carbonate in the original mixture.
3)Calculate the percentage of lithium carbonate in the original mixture.
4)Calculate the percentage of potassium carbonate in the original mixture.

Explanation / Answer

Since it is a mixture, the KOH and the K2CO3 react with the acid seperately:

(i) KOH(aq)+HCl(aq)?KCl(aq)+H2O(l)

(ii) 2K2CO3(aq)+4HCl(aq)
?4KCl(aq)+2H2O(l)+2CO2(g)

KCl does not react with the acid as it contains the same anion as the acid.

(a) We shall calculate the number of moles of CO2 obtained using the formula PV= nRT, where P is the pressure of CO2, V is the volume of CO2, n is the number of moles of CO2, R is the molar gas constant(8.31 J/mol/K) and T is the temperature of CO2 in Kelvin(K).

Let's convert all units to SI units:

1.00mL= 1.00x10^-6 m^3
249mL = (1.00x10^-6)x(249)
= 2.49x10^-4 m^3

760 torr= 1.01x10^5 N/m^2
740 torr= (((1.01x10^5)/760)x(740))
= 98342.10526 N/m^2

n= PV/RT
= (98342.10526)x(2.49x10^-4)/(8.31)
x(22+273)
= 9.988857293x10^-3

According to equation(ii), the ratio of the number of moles of K2CO3 that reacts to that of CO2 that forms is 1:1. Therefore, the number of moles of K2CO3 in the mixture
= number of moles of CO2 that forms
= 9.988857293x10^-3

Relative Formula Mass(RFM) of K2CO3
= 2(RAM of K)+(RAM of C)+3(RAM of O)
= 2(39.1)+12+16
= 106.2

1 mole of K2CO3= 106.2g
9.988857293x10^-3 moles of K2CO3
= (106.2)x(9.988857293x10^-3)
= 1.060816645g (mass of K2CO3 in the mixture)

Thus, the percentage of K2CO3 in the mixture
= (mass of K2CO3/mass of the mixture)x(100%)
= (1.060816645/5.00)x(100%)
= 21.2%

(b) HCl(aq)+NaOH(aq)?NaCl(aq)+H2O(l)

Number of moles of NaOH
= ([NaOH])x(volume of NaOH in L)
= (1.5)x(86.6x10^-3)
= 0.1299

According to the equation, the ratio of the number of moles of NaOH that reacts to that of HCl that reacts is 1:1. Therefore, the number of moles of HCl that reacted
= number of moles of excess HCl
= number of moles of NaOH
= 0.1299

According to equation (ii), the ratio of the number of moles of HCl that reacts to that of CO2 that forms is 2:1. Therefore, the number of moles of HCl that reacted with K2CO3
= 2(number of moles of CO2 formed)
= 2(9.988857293x10^-3)
= 0.019977714

Original number of moles of HCl
= ([HCl])x(volume of HCl solution)
= (2.00)x(0.100)
= 0.200

Number of moles of HCl that reacted with KOH
= original nHCl-(nHCl that reacted with K2CO3+nHCl that reacted with NaOH)
= 0.200-(0.019977714+0.1299)
= 0.050122286

Number of moles of KOH in the mixture
= number of moles of HCl that reacted with KOH
= 0.050122286

RFM of KOH= 56.1
0.050122286 moles of KOH
= (0.050122286)x(56.1)
= 2.811860245g (mass of KOH in the mixture)

Thus, percentage of KOH in the mixture
= (2.811860245/5.00)x(100%)
= 56.2%

Percentage of KCl in the mixture
= (100-(56.2+21.2))%
= 22.6%

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