Question

60cm3 of carbon (||) oxide,(co) are sparked with 30cm3 of oxygen if all the volumed of the gases are measured at s.t.p. calculate the volume of the residual gases after sparking​

Answer 1

We can begin by using the balanced chemical equation for the reaction between carbon monoxide and oxygen:

2CO + O2 -> 2CO2

This tells us that 2 volumes of CO react with 1 volume of O2 to produce 2 volumes of CO2. Since we have equal volumes of CO and O2, the limiting reactant will be the one that requires more volume, which is the CO:

2 volumes of CO + 1 volume of O2 -> 2 volumes of CO2

Using the ideal gas law, we can convert the volumes of CO and O2 at STP (standard temperature and pressure, which is 0°C and 1 atm) to moles:

n(CO) = V(CO) / Vm = 60 cm3 / 22.4 L/mol = 0.00268 mol

n(O2) = V(O2) / Vm = 30 cm3 / 22.4 L/mol = 0.00134 mol

Since 2 volumes of CO react with 1 volume of O2, we can say that the reaction will use up 2 x 0.00134 = 0.00268 mol of O2. Since we have 0.00134 mol of O2 initially, this means that all of the O2 will be used up in the reaction, leaving none in the residual gases. The reaction will also produce 2 x 0.00268 = 0.00536 mol of CO2.

Using the ideal gas law again, we can convert the volume of CO2 produced to a volume at STP:

V(CO2) = n(CO2) x Vm = 0.00536 mol x 22.4 L/mol = 0.120 cm3

Therefore, the volume of residual gases after the reaction is:

V(residual) = V(total) - V(CO) - V(O2) - V(CO2) = 90 cm3 - 60 cm3 - 30 cm3 - 0.120 cm3 = 0.880 cm3