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# The chief ore of Zn is the sulphide, ZnS. The ore is concentrated by froth floatation process and then heated in air to convert ZnS to ZnO. $2ZnS+3{{O}_{2}}\to 2ZnO+2S{{O}_{2}}$ (75% conversion)$ZnO+{{H}_{2}}S{{O}_{4}}\to ZnS{{O}_{4}}+{{H}_{2}}O$ (100% conversion)$2ZnS{{O}_{4}}+2{{H}_{2}}O\to 2Zn+2{{H}_{2}}S{{O}_{4}}+{{O}_{2}}$ (80% conversion)The mass of Zn obtained from the sample of ore containing 292.2 Kg of ZnS will be:[A] 117.72 Kg [B] 214.62 Kg[C] 145.23 Kg[D] 166.55 Kg

Last updated date: 14th Aug 2024
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Hint: To solve this, firstly calculate the number of moles of zinc in the given mass of ZnS. Then use it to find the moles of ZnS, $ZnS{{O}_{4}}$ and Zn considering the efficiency in each step. Do not forget to convert mass in Kg once you’ve found out the mass in the units of ‘g’.

We know that in metallurgy, we obtain pure metals through their ores. We purify ores through several methods and we obtain the required pure metal from the impure ore.
Now here to solve the given question, firstly let us calculate the number of moles of ZnS in the given mass i.e. 292.2Kg.

We know that, number of moles = $\dfrac{weight(g)}{molecular\text{ weight}}$
We have 292.2Kg of ZnS i.e. 292200g. And we know that the molecular weight of ZnS is 97.38 g/mol.
Therefore, putting the values in the above equation we will get that,
Number of moles = $\dfrac{292200\text{ }g}{97.4\text{ g/mol}}$ = 3000 moles of ZnS.
Now, from the first equation we can see that the ratio of ZnO produced from ZnS is 1:1 but the efficiency of the reaction is 75%.

So, moles of ZnO produces = number of moles of ZnS $\times$ efficiency of the reaction.
Therefore, the number of moles of ZnO produces = $3000\times \dfrac{75}{100}$ = 2250 moles.
Now from the second reaction we can see that ZnO is converted to $ZnS{{O}_{4}}$ and the efficiency is 100% so the number of moles of ZnO reacted will be equal to the number of moles of $ZnS{{O}_{4}}$ produced i.e. 2250 moles.
Now, from 2250 moles of $ZnS{{O}_{4}}$ we get zinc metal with 80% efficiency.
Therefore, number of moles of Zn metal = number of moles of $ZnS{{O}_{4}}$$\times$ efficiency of the reaction
So, the number of moles of Zn metal = $2250\times \dfrac{80}{100}$ = 1800 moles of Zn.
So, we can see from the above calculation that we obtained 1800 moles of Zn from 292.2 Kg of ZnS.

Now, mass of Zn produced = Number of moles of Zn $\times$ molar mass of Zn
We know the molar mass of Zn is 65.4 g/mol.
Therefore, mass of Zn produced = 1800 moles $\times$ 65.4 g/mol = 117720g.
Since, 1Kg = 1000g.
So, 117720 g = $\dfrac{117720}{1000}$ = 117.72 g.
So, 117.72g of Zn is produced from 292.2 kg of ZnS.
So, the correct answer is “Option A”.

Note: Zinc is obtained from its ore zinc blende which is also known as zinc sulphide. Zinc sulphide comes in two structures. One is sphalerite and the other is wurtzite. In both the forms zinc and sulphur ratio is 1:1 and maintains tetrahedral arrangement. However, Sphalerite is thermodynamically stable compared to wurtzite but both of them are found.