1.2g mixture of na2c03 and k2c03 was dissolved in water to form 100cm3 of solution, 20 cm3 of this solution required 40cm3 of 0.1 hcl for neutralization. Calculate the weight na2c03 and k2c03 in the mixture?
My calculation 20cm3/100cm3 = 0.2
x/53 + (0.2g-X) + y/69 = 0.004
0.01886g-0.01886x+ 0.01449x =0.004g
x = 0.000228/0.00437
x= 0.052/1.2 x 100
x = 4g k2c03
1g - 4g = 6g na2c03 Who check for me
To calculate the weights of Na2CO3 and K2CO3 in the mixture, you need to use the given information about the neutralization reaction between the solution and the 0.1 HCl.
First, find the number of moles of HCl used in the neutralization reaction:
Molarity (M) = moles/volume (L)
0.1 M HCl = x moles/0.04 L (40 cm3 = 0.04 L)
x = 0.004 moles
Since Na2CO3 and K2CO3 react with HCl in a 1:2 ratio, we know that the moles of Na2CO3 in the mixture are half of the moles of HCl:
Moles of Na2CO3 = 0.004 moles/2 = 0.002 moles
Next, calculate the molar masses of Na2CO3 and K2CO3:
Na2CO3: 2(22.99 g/mol) + 12.01 g/mol + 3(16.00 g/mol) = 105.99 g/mol
K2CO3: 2(39.10 g/mol) + 12.01 g/mol + 3(16.00 g/mol) = 138.21 g/mol
Now, use the molar masses to calculate the weights of Na2CO3 and K2CO3 in the mixture:
Weight of Na2CO3 = Moles of Na2CO3 x Molar mass of Na2CO3
= 0.002 moles x 105.99 g/mol
= 0.212 g
Weight of K2CO3 = Moles of K2CO3 x Molar mass of K2CO3
= 0.002 moles x 138.21 g/mol
= 0.276 g
Therefore, the weight of Na2CO3 in the mixture is 0.212 g, and the weight of K2CO3 is 0.276 g.
Let's go through the steps to calculate the weights of Na2CO3 and K2CO3 in the mixture:
1. Convert the volumes of the solution and the HCl used into moles:
- 20 cm3 solution * (1 L / 1000 cm3) = 0.02 L
- 40 cm3 0.1 M HCl * (1 L / 1000 cm3) = 0.004 mol HCl
2. Use the balanced chemical equation between Na2CO3 and HCl to determine the moles of Na2CO3:
- From the equation 2 Na2CO3 + 2 HCl ⟶ 2 NaCl + CO2 + H2O
- 2 mol Na2CO3 react with 2 mol HCl
- Therefore, 0.004 mol HCl would react with 0.004 mol Na2CO3
3. Calculate the molar mass of Na2CO3:
- Molar mass of Na2CO3 = (2 * atomic mass of Na) + atomic mass of C + (3 * atomic mass of O)
- ≈ (2 * 22.99 g/mol) + 12.01 g/mol + (3 * 16.00 g/mol)
- ≈ 46.00 g/mol + 12.01 g/mol + 48.00 g/mol
- ≈ 106.01 g/mol
4. Determine the weight of Na2CO3:
- 0.004 mol Na2CO3 * (106.01 g/mol) = 0.42404 g Na2CO3
5. Subtract the weight of Na2CO3 from the total weight of the mixture to find the weight of K2CO3:
- 1.2 g mixture - 0.42404 g Na2CO3 = 0.77596 g K2CO3
Therefore, according to the given calculation, the weight of Na2CO3 in the mixture is approximately 0.42404 g, and the weight of K2CO3 is approximately 0.77596 g.
I made a typo so I've erased my entire response and posted the correct answer.
You know your answer can't be right because you have 4 g K2CO3 but the entire sample is only 1.2 g and 4 g is more than the you started with
My calculation 20cm3/100cm3 = 0.2
Yes, 0.2 is the factor and the mass of the sample is 0.2*1.2 = 0.24g or another way to look at it is 1.2 x (20/100) = 0.24g titrated. However, you only need the factor.
x/53 + (0.2g-X) + y/69 = 0.004
This titration is for 0.2 of the original or 0.2x and 0.2y; therefore,
(0.2x/53) + (0.2y/69) = 0.004
Also note that where you have substituted 0.2-x for x it should be 1.2-y. Since x + y = 1.2 then x = 1.2-y
0.2x/53 + 0.2y/69 = 0.004
3.773E-3*x + 2.898E-3*y = 0.004
Substitute for x = 1.2-y
3.77E-3(1.2-y) + 2.898E-3*y = 0.004
I will let you finish this but the answer is 0.6 g Na2CO3 and 0.6g K2CO3. By the way, I used 69.1 for K2CO3 that will not affect the answer much.
0.01886g-0.01886x+ 0.01449x =0.004g
x = 0.000228/0.00437
x= 0.052/1.2 x 100
x = 4g k2c03
1g - 4g = 6g na2c03 Who check for me