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1.0 g of magnesium is burnt with 0.56 g `O_(2)` in a closed vessel. Which reactant is left in excess and how much?A. Mg,0.16 gB. `O_(2),0.16g`C. Mg,0.44 gD. `O_(2),0.28 g` |
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Answer» Correct Answer - A The balance chemical equation is `{:(Mg,+,(1)/(2)O_(2),to,MgO),(24g,,16g,,40g):}` From the above equation , it is clear that, 24 g of Mg reacts with 16 g of `O_(2)`. Thus , 1.0 g of Mg reacts with `(16)/(24)g` of `O_(2)=0.67g of O_(2)`. But only 0.56 g of ` O_(2)` is available which is less then 0.67 g . thus , `O_(2)` is the limiting reagent. Further , 16 g of `O_(2)` reacts with 24 g of Mg. `:.`0.56 g of `O_(2)` will with react with Mg `=(24)/(16)xx0.56=0.84g` `:.` Amount of Mg left unreacted =(1.0-0.84) g Mg =0. 16 g Mg |
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