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Current market prices of Al, Zn and Fe scraps per kg are Rs. 20, Rs. 16 and Rs. 3 respectively. If H_(2) is to be prepared by the reaction of one of these metals with H_(2)SO_(4), which would be the cheapest metal to use ? Which would be most expensive ? |
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Answer» Solution :The various chemical reactions INVOLVED are given below : (i) `underset(2xx27=54 G)(2Al)+3H_(2)SO_(4)rarr Al_(2)(SO_(4))_(3)+underset(3xx2=6 g)(3H_(2))` (ii) `underset(65 g)(Zn)+H_(2)SO rarr ZnSO_(4) + underset(2g)(H_(2))` (iii)`underset(56 g)(Fe)+H_(2)SO_(4)rarr FeSO_(4)+underset(2 g)(H_(2))` Step 1. To calculate the cost of preparation of 100 g of `H_(2)` from Al 6 g of `H_(2)` prepared from Al = 54 g `THEREFORE` 100 g of `H_(2)` will be obtained from `Al=(54)/(6)xx100=900 g` Cost of 1000 g of Al = Rs. 20 `therefore` Cost of 900 g of `Al=(20)/(1000)xx900=Rs. 18` Step 2. To calculate the cost of preparation of 100 g of `H_(2)` from Zn. 2 g of `H_(2)` is peoduced from Zn = 65 g `therefore` 100 g of `H_(2)` will be obtained from `Zn=(65)/(2)xx100=3250 g` Cost of 1000 g of Zn Rs. 16 `therefore` Cost of 3250 g of `Zn=(16)/(1000)xx3250=Rs.52`. Step 3. To calculate the cost of preparation of 100 g of `H_(2)` from Fe. 2 g of `H_(2)` is produced from Fe = 56 g `therefore` 100 of `H_(2)` will be obtained from `Fe=(56)/(2)xx100=2800 g` Cost of 1000 g of Fe = Rs. 3 `therefore` Cost of 2800 g of `Fe=(3)/(1000)xx2800=Rs. 8.40`. Thus, Fe is the cheapest and Zn is the most expensive metal to use for the preparation of `H_(2)`. |
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