When 8.21 l of c3h8 (g) burn in oxygen, how many liters of oxygen are consumed? all gas volumes are measured at the same temperature and pressure but not at stp?
8.21 L of C3H8(g)
Lets take c as the molar volume at that temperature. c L <><> 5c L C3H8 (g) + 5O2 (g) --> 3CO2 + 4H2O + Q 8.21 L <><> x L x = (8.21 * 5c)/c = 8.21 * 5 = 41.05 L O2 consumed for a 100% yield.
