28 gm lithium is mixed with 48 gm `O_(2)` to reacts according to the following reaction. `4Li + O_(2) to 2Li_(2)O` The mass of `Li_(2)O` formed is
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`4Li + O_(2) to 2Li_(2)O` moles taken `{:((28)/(7),(48)/(32)),(=4,=1.5):}` `("moles taken")/("stoich. Coeff") (4)/(4)=1" "(1.5)/(1)=1.5` `" "` (L.R.) Moles of `Li_(2)O` formed =`(2)/(4)x4=2` Mass of `Li_(2)O` formed=`2xx30=60` gm
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