10 g hydrogen is reacted with 64 g of oxygen. The amount of water formed will be (in moles)

Options

A. 3
B. 4
C. 1
D. 2

Answer

B

Explanation

`10gH_(2)=(10)/(2)"moles = 5 moles"`
`64g O_(2)=(64)/(32)" moles = 2 moles"`
`2H_(2)+O_(2)rarr 2H_(2)O`
2 moles of `H_(2)` react with 1 mole of `O_(2)`
`therefore " 5 moles of "H_(2)" all react with "O_(2)`
`=(1)/(2)xx5=2.5" moles"`
which is not present
1 mole of `O_(2)` reacts with `H_(2)=2` moles
`therefore` 2 moles of `O_(2)` will react with `H_(2)=4` moles
`H_(2)` left = 1 mole
Thus, `O_(2)` is the limiting reactant and `H_(2)` is excess reactant. `H_(2)O` formed from 2 moles of `O_(2)=4` moles.