Platinum (Pt) is used in part of your cars catalytic converter to reduce carbon
ID: 507939 • Letter: P
Question
Platinum (Pt) is used in part of your cars catalytic converter to reduce carbon monoxide emissions by converting carbon monoxide (CO) to carbon dioxide (CO2). This chemical reaction occurs on the surface of platinum, so rather than using a solid chunk of Pt, which is very expensive ($30-40 per gram), the catalyst consists of small, high surface area particles coated in Pt. The picture to the right depicts these platinum coated particles. The catalyst contains 6.6 mg Pt/g catalyst.
Using this platinum catalyst, you want to create a catalytic system for an automobile so it can handle the following requirements: A) When accelerating, the automobile engine produces 175 L/s of exhaust gases. CO is present at about 1.0 mg CO per L exhaust before passing through your catalyst. You want the catalytic converter to convert all of the CO(g) to CO2(g) before passing it through the rest of the exhaust system.
How many g of CO will pass through your catalyst per minute? I did this part. I got 10.5 g(CO)(1/min)
Q.
A) When operating properly, each gram of Pt can react 2.6 g of CO each minute.
How many g of Pt will you need?
How many g of catalyst will be necessary to contain this amount of Pt?
B) The catalyst has a density of 0.81 g/mL.
What volume of catalyst is needed to reduce CO emissions?
Thank you for your help!
Explanation / Answer
Ans. #1. Total exhaust per minute = emission per second x 60 seconds
= (175 L/ s) x 60s ; [1 min = 60s]
= 10500 L/ min
Mass of CO in total exhaust in 1 minute = CO content of exhaust x (Exhaust/ min)
= (1.0 mg / L) x 10500 L
= 10500 mg
= 10.500 g ; [1 g = 1000 mg]
Therefore, CO exhaust rate = 10.500 g/ min
#A. 1.0 g Pt reacts with 2.6 g CO per minute. The engine emits 10.5 g CO per minute.
Thus, on a per minute basis-
2.6 g CO is reacted by 1.0 g Pt
Or, 1.0 g - - (1.0 / 2.6) g Pt
Or. 10.50 g - - (1.0 / 2.6) x 10.50 g Pt
= 4.04 g
Thus, required mass of Pt = 4.04 g
#B. Volume of Pt needed = Required mass of Pt / Density of Pt
= 4.04 g / (0.81 g/ mL)
= 4.98 mL