Question

A sample of cacl2⋅2h2o/k2c2o4⋅h2o solid salt mixture is dissolved in ~150 ml de-ionized h2o. the oven dried precipitate has a mass of 0.333 g. the limiting reactant in the salt mixture is k2c2o4⋅h2o. cacl2⋅2h2o(aq) + k2c2o4⋅h2o(aq) à cac2o4⋅h2o(s) + 2kcl(aq) + 2h2o(l) starting material (sm) product molar mass (mm) g/mol: cacl2⋅2h2o = 147.02 k2c2o4⋅h2o = 184.24 cac2o4 = 128.10 determine mass of k2c2o4⋅h2o(aq) in salt mixture in grams. answer to 3 places after the decimal and include unit, g

Answer 1

We are given that the balanced chemical reaction is:

cacl2⋅2h2o(aq) + k2c2o4⋅h2o(aq) ---> cac2o4⋅h2o(s) + 2kcl(aq) + 2h2o(l)

We known that the product was oven dried, therefore the mass of 0.333 g pertains only to that of the substance cac2o4⋅h2o(s). So what we will do first is to convert this into moles by dividing the mass with the molar mass. The molar mass of cac2o4⋅h2o(s) is molar mass of cac2o4 plus the molar mass of h2o.

molar mass cac2o4⋅h2o(s) = 128.10 + 18 = 146.10 g /mole

moles cac2o4⋅h2o(s) = 0.333 / 146.10 = 2.28 x 10^-3 moles

Looking at the balanced chemical reaction, the ratio of cac2o4⋅h2o(s) and k2c2o4⋅h2o(aq) is 1:1, therefore:

moles k2c2o4⋅h2o(aq) = 2.28 x 10^-3 moles

Converting this to mass:

mass k2c2o4⋅h2o(aq) = 2.28 x 10^-3 moles (184.24 g /mol) = 0.419931006 g

 

Therefore:

The mass of k2c2o4⋅h2o(aq) in the salt mixture is about 0.420 g

Answer 2

Answer: The mass of $K_2C_2O_4.H_2O$ in the salt mixture is 0.424 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$      .....(1)

• For $CaC_2O_4.H_2O$ :

Given mass of $CaC_2O_4.H_2O$ = 0.333 g

Molar mass of $CaC_2O_4.H_2O$ = 146.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of}CaC_2O_4.H_2O=\frac{0.333g}{146.12g/mol}=0.0023mol$

The given chemical equation follows:

$CaCl_2.2H_2O(aq.)+K_2C_2O_4.H_2O(aq.)\rightarrow CaC_2O_4.H_2O(s)+2KCl(aq.)+2H_2O(l)$

By Stoichiometry of the reaction:

1 mole of $CaC_2O_4.H_2O$ is produced by 1 mole of $K_2C_2O_4.H_2O$

So, 0.0023 moles of $CaC_2O_4.H_2O$ will be produced by = $\frac{1}{1}\times 0.0023=0.0023mol$ of $K_2C_2O_4.H_2O$

Now, calculating the mass of $K_2C_2O_4.H_2O$ by using equation 1, we get:

Molar mass of $K_2C_2O_4.H_2O$ = 184.24 g/mol

Moles of $K_2C_2O_4.H_2O$ = 0.0023 moles

Putting values in equation 1, we get:

$0.0023mol=\frac{\text{Mass of }K_2C_2O_4.H_2O}{184.24g/mol}\\\\\text{Mass of }K_2C_2O_4.H_2O=(0.0023mol\times 184.24g/mol)=0.424g$

Hence, the mass of $K_2C_2O_4.H_2O$ in the salt mixture is 0.424 grams.