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How many gram of oxygen `(O_(2))` is required to completely react with 0.200 g of hydrogen `(H_(2))` to yield water `(H_(2)O)` ? Also calculate the amount of water formed (molecular mass H = 2, O = 32).

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The balanced equation for the reaction is
`underset(4g)underset("2 mol")(2H_(2))+underset(32 g)underset("1 mol")(O_(2))rarr underset(36 g)underset("2 mol")(2H_(2)O)`
Now, 4 g of `H_(2)` require oxygen = 32 g
0.200 g of `H_(2)` require oxygen `=(32)/(4)xx0.200=1.6 g`
Again, 4 g of `H_(2)` produce `H_(2)O=36 g`
0.200 g of `H_(2)` produce `H_(2)O=(36)/(4)xx0.200=1.8g`.

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