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2g of `H_(2) and 1g O_(2)` are allowed to react according to following equation
`2H_(2)(g)+O_(2)(g) to 2H_(2)O(g)`
Amount of `H_(2)O` formed in the reaction will be:
A. 3g
B. 1.125g
C. 4.5g
D. 2.50g

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