How much amount of water will be produced when 10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded also identify the limiting reagent?

Text Solution

1 mol2mol 3 mol4 mol

Answer : D

Solution : Moles of hydrogen `=(10)/(2)=5`mol <br> Moles of oxygen `=(64)/(32)=2` mol <br> `2H_(2)+O_(2)to2H_(2)O` <br> Thus, `O_(2)` is the limiting reactant <br> `:.` amount of water produced `=2xx2`mol <br> 4 mol <br> `:.` molarity `=(1.096)/(0.0833L)` <br> `13.15` mol L or 13.15M

Text Solution

2 mol3 mol4 mol3 mol

Answer : C

Solution : `underset(2mol)(2H_(2)(g)) + underset(1mol)(O_(2)(g)) rarr underset(2mol)(2H_(2) O(l))` <br> `n_(H_(2)) = (10g)/(2g mol^(-1)) = 5 mol` <br>`n_(O_(3)) = (64g)/(32g mol^(-1)) = 2 mol` <br> Moles of product formed from each rectant: <br> `5 mol H_(2) xx (2 mol H_(2) O)/(2 mol H_(2)) = 5 mol H_(2) O` <br> `2 mol O_(2) xx (2 mol H_(2)O)/(1 mol O_(2) = 4 mol H_(2) O` <br> Thus, `O_(2)` is the limiting reacant and `4 mol H_(2) O` will be produced.

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10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. What will be the amount of water produced in this reaction?A. 1 mol B. 2 molC. 3 molD. 4 mol