Chemistry Question: I2O5(g) + 5 CO(g) → 5 CO2(g) + I2(s) ?
Please answer and explain the following:
Consider the reaction:
I2O5(g) + 5 CO(g) → 5 CO2(g) + I2(s)
If 80.0g of iodine(V) oxide, reacts with 28.0g of carbon monoxide.
a) Determine the limiting reagent
b) What mass of iodine can be produced in the reaction?
c) What is the percentage yield if the mass of iodine formed in the reaction is 40.6g?
d) What volume of CO2(g) would also be produced in the reaction at 35oC and 95kPa of pressure?
Thank you!!
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Verified answer
n(I2O5)=0.2397 (80/333.8) R=8.315
n(CO)=0.9996 (28/28.01) K= C + 273.1
Note how I take the limiting regents for the stoichiometric ratio.
a) n(CO) = 5n(I2O5) -from equation
= 5*0.2397 = 1.1985
But n(CO) was given to be exactly 0.9996, hence it is in excess by (1.1985-0.9996) meaning I205 is the limiting reagent
b) Taking the limiting reagent I205:
n(I2) = n(I2O5) -from equation
= 0.2397
m(I2) = nM = 0.2397*253.8
= 60.84g
c) (40.6/60.84) * 100 = 66.73%
d) n(CO2) = 5n(I205)
= 5*0.2397
= 1.1985
V(CO2) = (n*R*T) / P
= [1.1985*8.315*(35+273.1)] / 95
=32.32L
1. calculate the mol. wt of I2O5, see 80 gms are how many moles=x.
2. 1 mole CO=28 gm= what we are using. So limit of the yield will be the no. of moles of I2O5=no. of moles of I2 =x.
3.yield of iodine[ mol wt=129*2] = 40.6/129 *2moles of I2 = y ; so % = pl. compute ! .As we are using surplus CO, if loss = 0, yield % has to be 100 !
4. y moles of CO2 is formed , you can calculate the volume it occupies at any temp/ pressure using PV/T = P'V"/T'., here T is in abs degrees !