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You are given a bottle of impure CoSO4 7H2O which is thought to be 74.4% by weight CoSO4 7H2O. You propose to run your own analysis to find out. What weight of sample should you take to get 0.345 g of BaSO4?

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sebassandin

Answer:

[tex]m_{CoSO_4\ \ 7H_2O}^{sample}=0.558gm_{CoSO_4\ \ 7H_2O}[/tex]

Explanation:

Hello.

In this case, considering a representative reaction by which BaSO4 is yielded via CoSO4, for instance:

[tex]CoSO_4\ \ 7H_2O+BaCl_2\rightarrow BaSO_4+CoCl_2+7H_2O[/tex]

Since it is desired to obtain 0.345 g of BaSO4 (molar mass = 233.38 g/mol), we can compute the theoretical mass of CoSO4·7H20 (molar mass = 281 g/mol) via the following stoichiometric procedure:

[tex]m_{CoSO_4\ \ 7H_2O}=0.345gBaSO_4*\frac{1molBaSO_4}{233.38 gBaSO_4}*\frac{1molCoSO_4\ \ 7H_2O}{1molBaSO_4} *\frac{281.00gCoSO_4\ \ 7H_2O}{1molCoSO_4\ \ 7H_2O} \\\\m_{CoSO_4\ \ 7H_2O}=0.415gCoSO_4\ \ 7H_2O[/tex]

Next, given the percent purity of the cobalt (II) sulfate heptahydrate, we can compute the mass of the whole sample as shown below:

[tex]m_{CoSO_4\ \ 7H_2O}^{sample}=0.415gCoSO_4\ \ 7H_2O^{pure}*\frac{100gCoSO_4\ \ 7H_2O^{sample}}{74.4gCoSO_4\ \ 7H_2O^{pure}} \\\\m_{CoSO_4\ \ 7H_2O}^{sample}=0.558gm_{CoSO_4\ \ 7H_2O}[/tex]

Best regards!

dsdrajlin

To obtain 0.345 g of BaSO₄, we need to weigh 0.559 g of CoSO₄.7H₂O with a purity of 74.4%.

Let's consider a possible reaction for the obtaining of BaSO₄ from CoSO₄.7H₂O.

CoSO₄.7H₂O(aq) + Ba(NO₃)₂(aq) ⇒ BaSO₄(s) + Co(NO₃)₂(aq) + 7 H₂O(l)

We want to find the mass of CoSO₄.7H₂O required to form 0.345 g of BaSO₄. We will consider the following relationships:

  • The molar mass of BaSO₄ is 233.38 g/mol.
  • The molar ratio of BaSO₄ to CoSO₄.7H₂O is 1:1.
  • The molar mass of 281.10 g/mol.

[tex]0.345gBaSO_4 \times \frac{1molBaSO_4}{233.38gBaSO_4} \times \frac{1molCoSO_4.7H_2O}{1molBaSO_4} \times \frac{281.10gCoSO_4.7H_2O}{1molCoSO_4.7H_2O} =0.416gCoSO_4.7H_2O[/tex]

The purity of the sample is supposed to be 74.4%, that is, there are 74.4 g of CoSO₄.7H₂O every 100 g of sample. The mass of the sample that contains 0.416 g of CoSO₄.7H₂O is:

[tex]0.416gPure \times \frac{100gSample}{74.4gPure} = 0.559gSample[/tex]

To obtain 0.345 g of BaSO₄, we need to weigh 0.559 g of CoSO₄.7H₂O with a purity of 74.4%.

Learn more: https://brainly.com/question/8684258

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