1.

How many gm of bromine will react with 21 gm C_3H_6

Answer»

80
160
240
320

Solution :`undersetundersetunderset"42 GMS""1 MOLE""PROPENE"(CH_3-CH)=CH_2+undersetunderset"160 gms""1 mole"(Br_2)to CH_3-underset"1,2-Dibromo propane"(undersetunderset(Br)(|)CH_2-undersetunderset(Br)(|)CH_2)`
`because` 24 gms of propene reacts with 160 gms of bromine
`because` 21 gms of propene `160/42xx21=80` gms


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