Asample 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

The mass of in the salt mixture is 0.424 grams.

Explanation:

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

     .....(1)

Given mass of = 0.333 g

Molar mass of = 146.12 g/mol

Putting values in equation 1, we get:

The given chemical equation follows:

By Stoichiometry of the reaction:

1 mole of is produced by 1 mole of

So, 0.0023 moles of will be produced by = of

Now, calculating the mass of by using equation 1, we get:

Molar mass of = 184.24 g/mol

Moles of = 0.0023 moles

Putting values in equation 1, we get:

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

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i believe the answer is three water molecules