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3.0 g of H(2) react with 29.0 g of O(2)...

3.0 g of `H_(2)` react with 29.0 g of `O_(2)` yield `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 reactant left unreacted

Text Solution

Verified by Experts

`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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