First we need to write a balanced chemical equation:

a #CS_2# + b #O_2 -># c #CO_2# + d #SO_2#

If we set a = 1, c is also 1, and d is 2, since each mole of #CS_2# will produce one mole of #CO_2# and two moles of #SO_2#

#CS_2# + b #O_2 -># #CO_2# + 2 #SO_2#

Now there are 6 moles of #O# on the right, so we need 3 #O_2# on the left to balance the equation:

#CS_2# + 3 #O_2 -># #CO_2# + 2 #SO_2#

Now, the molar mass of #CS_2# is 76 g/mol. We find the number of moles present:

#n=m/M = 7.6/76 = 0.1# #mol#

The balanced equation tells us that #0.1# #mol# of #CS_2# will produce #0.1# #mol# of #CO_2# and #0.2# #mol# of #SO_2#

#m_(CO2) = nM = 0.1 xx 44= 4.4# #g#

#m_(SO2) = nM = 0.2 xx 64= 12.8# #g#