Question

given the following questions, i would like to ... Question Given the following questions, I would like to know the answers to c, d, and e. located at the bottom (due to 4 answers per post limit). Thanks! Platinum (Pt) is used in part of your car's catalytic converter to reduce carbon monoxide emissions by the following chemical reaction: 2CO(g) + O2(g) Pt 2CO2(g) ---> 1) A matrix contains 7.1mg Pt/g of matrix. What is the %Pt in the matrix by mass? mass of pt = 7.1 mg mass of matrix = 1 g = 1000 mg mass % of Pt = mass of Pt *100 /mass of matrix = 7.1*100/1000 =0.71 % Answer: 0.71 % 2) 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 148 L/s of exhaust gases. CO is present at about 2 mg CO per 1 L exhaust before passing through your catalyst matrix. 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 grams of CO will pass through your catalytic matrix per minute? 1 minute = 60 s volume of exhaust gas produced in 1 min = 60 * 148 L = 8880 L amount of CO present in 1 L = 2 mg Total amount of CO present = 2*8880 L = 17760 mg = 17.76 g Answer: 17.76 g B) When operating properly, each gram of Pt can react 2.6 g of CO each minute. How many g of Pt catalyst will you need? How many g of matrix will be necessary to contain this amount of Pt? Mass od CO = 17.76 g 1 g of Pt reacts with 2.6 g CO mass of Pt required = 17.76 g /2.6 g = 6.83 g <----Answer 1 g of matrix has 7.1 mg = 0.0071 g of Pt So, for 6.83 g of Pt, matrix required = 6.83/0.0071 g = 962 g <---Answer c. The matrix has a high surface area (240 m2/g matrix) to allow exhaust gases to contact as much Pt as possible. How much surface area will this much catalyst matrix have? d. The catalyst matrix has a density of 0.81 g/mL. What volume of catalytic matrix is needed to reduce CO emissions? e. What if you were designing an engine for a SUV that produced 266 L/s of exhaust with a CO content of 3.4 mg CO per 1L of exhaust? What volume of catalytic matrix would you need to reduce the CO emission in this automobile?

Answer 1

c. The matrix has a high surface area (240 m2/g matrix) to allow exhaust gases to contact as much Pt as possible. How much surface area will this much catalyst matrix have?

Answer= The matrix contains 7.1mg Pt/g of matrix.

So, Surface area of catalyst matrix=240m2/7.1 mg pt=33.80 m2/mg catalyst (answer)

d. The catalyst matrix has a density of 0.81 g/mL. What volume of catalytic matrix is needed to reduce CO emissions?

Density of catalyst=0.81g/mL

Given

When operating properly, each gram of Pt can react 2.6 g of CO each minute.

Total amount of CO present = 17.76 g

So, amount of pt required=17.76g/2.6g=6.83 g pt

0.81 g pt=1ml

Gives, 1g pt=1/0.81 ml

So, 6.83 g pt=6.83/0.81 ml=8.43 ml( answer)

e. What if you were designing an engine for a SUV that produced 266 L/s of exhaust with a CO content of 3.4 mg CO per 1L of exhaust? What volume of catalytic matrix would you need to reduce the CO emission in this automobile?

1 minute = 60 s

volume of exhaust gas produced in 1 min = 60 * 266 L = 15960 L

Amount of CO present in 1 L = 3.4 mg

Total amount of CO present = 3.4 mg*15960 L = 54264 mg = 54.264 g (Answer)

(As gram of Pt can react 2.6 g of CO each minute

So, Mass of catalyst matrix required=54.264 g pt /2.6 g of CO each minute=1132.53 g

Volume of catalyst matrix required=1132.53/0.81=1398.19 ml(answer) (Given, 1g pt=1/0.81 ml)