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How much Cu can be obtained from 100g of $CuS{{O}_{4}}$ ?

Last updated date: 20th Jun 2024
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Hint: One mole of Copper Sulphate is generated from one mole of $C{{u}^{2+}}$ and a mole of $S{{O}_{4}}^{2-}$ ions. The molar mass of $CuS{{O}_{4}}$ and Cu are 160 grams and 64 grams respectively.

Step-by-Step Solution:
Before moving onto the specifics of this question, let us first look at how the compound $CuS{{O}_{4}}$ is made and the chemical reaction involved along with its equation.
- Copper sulphate is produced industrially by treating copper metal with hot concentrated sulfuric acid or its oxides with dilute sulfuric acid.
- It can be noted that the oxidation state exhibited by the copper atom in a $CuS{{O}_{4}}$ molecule is +2.
The ionic reaction of the above reaction is given by:
\[C{{u}^{2+}}+S{{O}_{4}}^{2-}\to CuS{{O}_{4}}\]
Having now understood how $CuS{{O}_{4}}$ is prepared, let us now move on to the specifics of this question.
Molecular mass of $CuS{{O}_{4}}$ = Atomic mass of Cu + Atomic mass of S + 4(Atomic mass of O)
Molecular mass of $CuS{{O}_{4}}$ = 64 + 32 + 4(16)
Molecular mass of $CuS{{O}_{4}}$ = 160$gmmo{{l}^{-1}}$
Mass of $CuS{{O}_{4}}$ given= 100 g
\[\text{Moles of CuS}{{\text{O}}_{4}}=\dfrac{\text{Weight of CuS}{{\text{O}}_{4}}}{\text{Molecular weight of CuS}{{\text{O}}_{4}}}\]
\[\text{Moles of CuS}{{\text{O}}_{4}}=\dfrac{100gm}{160gm}\]
\[\text{Moles of CuS}{{\text{O}}_{4}}=0.625mole\]
- We have already established above that one mole of $CuS{{O}_{4}}$ contains one mole of $C{{u}^{2+}}$ and one mole of $S{{O}_{4}}^{2-}$ .
So, we can say that 0.625 mole of $CuS{{O}_{4}}$ contains 0.625 moles of $C{{u}^{2+}}$ and 0.625 moles of $S{{O}_{4}}^{2-}$ .
Thus, \[Mass\text{ }of~C{{u}^{2+}}in\text{ }100\text{ }grams\text{ }of~CuS{{O}_{4}}=\text{ Moles of copper }\times \text{ Atomic weight of Copper}\]
\[Mass\text{ }of~C{{u}^{2+}}in\text{ }100\text{ }grams\text{ }of~CuS{{O}_{4}}=\text{ 0}\text{.625}\times 64\]
\[Mass\text{ }of~C{{u}^{2+}}in\text{ }100\text{ }grams\text{ }of~CuS{{O}_{4}}=\text{ 40gm}\]

Hence, by our calculations, we can safely conclude that 40 grams of Copper is obtained from 100 grams of $CuS{{O}_{4}}$ solution.

Note: This solution is directly related to the law of constant proportions which states that a given chemical compound always contains its component elements in a fixed ratio and does not depend on its source and method of preparation.