Question #93cb7

1 Answer
Apr 20, 2016

Answer:

#H_2# is the limiting reactant.
10 moles of #HBr# are produced.
160 g of #Br_2# are remaining after reaction.

Explanation:

Let us write again the balanced equation:

#H_2(g)+Br_2(g)->2HBr(g)#

In order to find the amount of #HBr(g)# that will form we will need to know which one is the limiting reactant.

#color(green)("Using 5mol "H_2)#:

#?molHBr=5cancel(molH_2)xx(2molHBr)/(1cancel(molH_2))=color(green)(10molHBr)#

#color(red)("Using 7mol "Br_2#:

#?molHBr=7cancel(molBr_2)xx(2molHBr)/(1cancel(molBr_2))=color(red)(14molHBr)#

In general, the reactant that produces the least amount of #HBr# is the limiting reactant. Therefore, #H_2# is the #color(green)("limiting reactant")# and #Br_2# is the #color(red)("excess reactant")#.

Since #color(green)(1molH_2)# reacts with #color(red)(1molBr_2)#, therefore, #color(green)(5molH_2)# will react only with #color(red)(5molBr_2)# and the excess amount of #Br_2# is #color(red)(2molBr_2)#.

In order to find the mass in grams for the excess #Br_2# remaining after reaction:

#n=m/(MM)=>m=nxxMM#

#m=2cancel(molBr_2)xx(80gBr_2)/(1cancel(molBr_2))=160gBr_2#