Answer:
7.6 mol Ar
0.44 mol C2H6
STP = 1 atm and 273 K
Use PV = nRT and solve for V
V = nRT/P
Plug in the number of moles accordingly for each gas and plug in 273 K for the variable T
1 atm for the variable P and use 0.08206 L•atm/k•mol for the variable R (gas law constant)