1.

3.0 g of H_(2) react with 29.0 g of O_(2) to form H_(2)O. (i) which is the limiting reagent? (ii) Calculate the maximum amount of H_(2)O that can be formed. (iii) Calculate the amount of the reactant left unreacted. Molecular mass of H_(2)=2.016.

Answer»

Solution :`UNDERSET(2xx2.016=4.032g)(2H_(2))+underset(32g)(O_(2))rarrunderset(2XX(1.016+16)=36.032)(2H_(2)O)`
`"3 g of "H_(2)" require "O_(2)=(32)/(4.032)xx3=23.8g`
Thus, `O_(2)(29g)` is present Hence, `H_(2)` is the limiting reactant.
`H_(2)O" FORMED "=(36.032)/(4.032)xx3g=26.8g`
`O_(2)" left unreacted "=29-23.8=5.2g`


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