0.50g of a mixture of K2c03 and Li2c03 requires 30ml of a 0.25N hcl solution for neutralization. What is the percentage composition of the mixture?
The answer k2c03 96%
Li2c03 4%
K2c03 weight 138g/mol
138/2 = 69
Li2c03 weight 74g/mol
74/2 = 37
30ml x 0.25N = 7.5 / 1000
x= 0.0075
Who helps me to solve it for me
Please Bob
You have two unknowns; therefore, you must have two equations in the two unknowns.
Let X = mass K2CO3
and Y = mass Li2CO3
---------------------
eqn 1:
X + Y = 0.500g
eqn 2:
(#Eq X) + (#Eq Y) = 0.0075 or
(X/69) + (Y/37) = 0.0075
Solve eqn 1 and 2 simultaneously for X an Y. Then
(X/0.5)*100 = %K2CO3
(Y/0.5)&100 = %Li2CO3
To solve this problem, you need to use the concept of neutralization reactions and stoichiometry. The steps to solve it are as follows:
1. Determine the number of moles of HCl used in the reaction.
- Given that the volume of the 0.25N HCl solution is 30 mL, we need to convert it to moles using the formula Molarity (M) = Moles (mol) / Volume (L).
- Convert 30 mL to L: 30 mL = 30/1000 = 0.03 L
- Molarity of HCl = 0.25N = 0.25 mol/L
- Moles of HCl = Molarity x Volume = 0.25 x 0.03 = 0.0075 moles
2. Determine the molar ratio between HCl and the carbonates.
- The balanced equation for the neutralization reaction between HCl and carbonates is:
HCl + 2MOH → MCl2 + H2O (M represents either K or Li)
- Since the molar ratio between HCl and the carbonates is 1:2, we can conclude that the moles of carbonates in the mixture are twice the moles of HCl used.
3. Determine the moles of the carbonate mixture.
- Moles of the carbonate mixture = 2 x Moles of HCl = 2 x 0.0075 = 0.015 moles
4. Determine the individual masses of K2CO3 and Li2CO3 in the mixture.
- Given the molar masses:
- K2CO3 = 2(39.10 g/mol) + 12.01 g/mol + 3(16.00 g/mol) = 138.21 g/mol
- Li2CO3 = 2(6.94 g/mol) + 12.01 g/mol + 3(16.00 g/mol) = 73.89 g/mol
- Now we can calculate the masses of K2CO3 and Li2CO3 using their respective molar masses:
- Mass of K2CO3 = Moles of K2CO3 x Molar mass of K2CO3 = 0.015 moles x 138.21 g/mol = 2.073 g
- Mass of Li2CO3 = Moles of Li2CO3 x Molar mass of Li2CO3 = 0.015 moles x 73.89 g/mol = 1.109 g
5. Determine the percentage composition of the mixture.
- Percentage of K2CO3 = (Mass of K2CO3 / Mass of mixture) x 100
= (2.073 g / (2.073 g + 1.109 g)) x 100 = 65.17%
- Percentage of Li2CO3 = (Mass of Li2CO3 / Mass of mixture) x 100
= (1.109 g / (2.073 g + 1.109 g)) x 100 = 34.83%
Therefore, the percentage composition of the mixture is approximately:
- K2CO3: 65.17%
- Li2CO3: 34.83%