The number of grams of bromine that will comp | vedclass

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Question

(A) The chemical reaction is: $C_{5}H_{10} + Br_{2} \rightarrow C_{5}H_{10}Br_{2}$From the stoichiometry,$1 \ mol$ of pent$-1-$ene reacts with $1 \ mo

Vedclass · Accepted Answer

$1143$

Vedclass · Answer

(A) The chemical reaction is: $C_{5}H_{10} + Br_{2} ightarrow C_{5}H_{10}Br_{2}$From the stoichiometry,$1 \ mol$ of pent$-1-$ene reacts with $1 \ mol$ of $Br_{2}$.Molar mass of pent$-1-$ene $(C_{5}H_{10})$ $= 5 	imes 12 + 10 	imes 1 = 70 \ g/mol$.Molar mass of $Br_{2} = 2 	imes 80 = 160 \ g/mol$.Moles of pent$-1-$ene $= \frac{5.0 \ g}{70 \ g/mol} = \frac{1}{14} \ mol$.Since the molar ratio is $1:1$,moles of $Br_{2}$ required $= \frac{1}{14} \ mol$.Mass of $Br_{2}$ required $= 	ext{moles} 	imes 	ext{molar mass} = \frac{1}{14} 	imes 160 \ g = 11.4285 \ g$.Expressing in terms of $10^{-2} \ g$: $11.4285 	imes 100 	imes 10^{-2} \ g = 1142.85 	imes 10^{-2} \ g$.Rounding to the nearest integer,we get $1143 	imes 10^{-2} \ g$.

The number of grams of bromine that will completely react with $5.0 \ g$ of pent$-1-$ene is $...... \times 10^{-2} \ g$. (Atomic mass of $Br = 80 \ g/mol$) [Nearest Integer]

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