Question

For the reaction CS2(l) + 3O2(g) --> CO2(g) + 2SO2(g) What volume of O2(g) is required to react with excess CS2(l) to produce 10.0L of CO2(g)? (Assume all gases are measured at 0°C and 1atm.)

a)30L

b)10L

c) 7.47L

d)44.8L

e)224L

Answer 1

CS2(l) + 3O2(g) --> CO2(g) + 2SO2(g)

The reaction suggest , 1 mole of CS2 reacts with 3 mole of Oxygen to produce 1 mole of CO2 and 2 moles of SO2

since volume of CS2 is excess, oxygen will be the limiting reactants and therefore

3 moles of O2 gives reacts with excess CS2 to give 1 mole of CO2.

from gas law PV= nRT, n= no of moles = PV/RT, P= pressrue, V= volume and R = gas constant and R= 0.0821 L.atm/mole.K, T = temperature

since pressrue and temperature remains constant during the reaction, V= Constant* n , volume is proportional ot no of moles

10 mole of CO2 correspond to 1 mole of CO2. 3 moles of O2 correspond to s 3*10= 30L of O2 ( a is correct)