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#
