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3-0 g of H2 react with 29-0 g of O2 yield H20.(i) which is the limiting reactant ?(ii) Calculate the maximum amount of H20 that(iii) Calculate the amount of one of the reactants which remain unreacted.can be formed.[Ans. (i) H2 (ii) 150 m |
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Answer» The balanced equation for the above reaction is as follows: 2H2 + O2 = 2H2O From the above equation it is clear that 2 mole H2 react with 1 mole O2 Molar mass of H2 = 2g Molar mass of O2= 32 g This implies, 4 g H2 react with 32 g O2 3 g H2 reacts with = (32/4) x 3g of O2 gas= 24 gAs the given amount of O2 is more than required therefore O2 is excess reagent and H2 is limiting reagent. 2 mole of hydrogen gas reacts to form 2 mole of water molecule therefore,4 g of H2 produces = 36 g of water So the amount of H2O produced by 3 g H2 = (36/4) x 3= 27 gHence, 27 g of water will be produced during the recation As, 24 g of oxygen has been utilised during the recation and 29 g of oxygen was supplied therefore amount of oxygen gas left is (29-24) = 5g |
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