When 7.25 g of #Al# reacts with excess #H_2SO_4# in the reaction #2Al (s) + 3 H_2SO_4 (aq) -> Al_2(SO_4)_3(aq) + 3H_2(g)# how many liters of #H_2# is formed at 25° C and a pressure of 1.650 atm?

1 Answer
Mar 10, 2016

Answer:

The first step is to calculate the number of moles of #Al# present, #n=0.27# #mol#, then use the mole ratio in the equation to determine that this releases #0.40# #mol# of #H_2#, with a volume of #9.0# #L# at STP. Next convert to the given temperature and pressure, and #V=5.95# #L#.

Explanation:

#2# #Al(s)# + #3# #H_2SO_4(aq) to # #Al_2(SO_4)_3(aq)# + #3# #H_2(g)#

First calculate the number of moles of #Al# in #7.25# #g#, knowing the molar mass of #Al# is #27# #gmol^-1#:

#n=m/M=7.25/27=0.27# #mol#

From the equation we can see that each #2# #mol# of #Al# yields #3# #mol# of #H_2#, so multiply this by #3/2# to find the number of moles of #H_2# released.

#n=0.40# #mol#

We know that a mole of an ideal gas occupies #22.4# #L# at standard temperature and pressure (STP), which is #0^o# #C# or #273# #K# and #1# #atm#, so this number of moles will produce #0.4xx22.4=9.0# #L# of gas at STP.

We are not asked about STP, though, but #25^o# #C# (#273# #K#) and #1.65# #atm#.

Using the combined gas law:

#(P_1V_1)/T_1=(P_2V_2)/T_2#

Rearranging to make #V_2# the subject:

#V_2=(P_1V_1)T_2/T_1P_2=(1*9*298)/(273*1.65)=5.95# #L#