how much carbon disulphide will be formed if 4 gram of carbon was heated with 8 gram of sulphur also identify the limiting reagent
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Explanation:
B. 79.2%
C+2S→CS2
Ideally, 1 mole of C will react with 2 moles of S to form CS2
1 mol of C = 12g 1 mol of S = 32g
12g carbon reacts with = 64g of Sulphur 1g carbon reacts with = 6412g of Sulphur
4g carbon reacts with = 6412×4=643g of Sulphur
But we have only 8g of sulphur
It mean sulphur is limiting agent.
due to sulphur is limiting reagent reaction will proceed according to sulphur .
64g sulphur formed = 76g CS2
1g S formed= 7664 g CS2
8g S formed = 7664×8=9.5g CS2
Hence, % purity =Wt. of productWt. of reactants×100 = 9.5(4+8)×100=9.512×100
= 79.16 % = 79.2 %
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