Carbon monoxide (CO) and hydrogen (H2) are fed to a continuous catalytic reactor operating at steady state. There are no other components in the feed. The outlet stream contains unconverted CO and H2, along with the products methanol (CH30H), ethanol (C2HsOH), isopropanol (C3H70H), and carbon dioxide (C02). These are the only species in the product stream. The reactions occuring are:

CO+ 2H2 → CH3OH
3CO+ 3H2 → C2H5OH+ CO2
5CO+ 4H2 → C3H7OH + 2CO2

The feed rates of CO and H2 to the reactor are 100 mol/h(each). the rates in the stream that leaves the reactor (in mols/h) are H2-30; CO-30; C2H5OH-5. What’s the mole fraction of each species in the product steam?

Explanation:

Let the products in the outlet streams be P(i.e. CH₃OH), Q(i.e. C₂H₅OH) and R(i.e. C₃H₇OH) respectively

Then the mass balance for CO, H and C₂H₅OH can be computed as follows:

For CO, The mass balance is;

100 - P - 3Q - 5R = 30    --- (1)

For H₂, The mass balance is;

100 - 2P - 3Q - 4R = 30    --- (2)

For C₂H₅OH, The mass balance is;

Q = 5    --- (3)

Replacing the value of equation(3) into equation (1) and (2); we have:

From equation (1):

100 - P - 3(5) - 5R = 30    --- (1)

100 - 15 - P - 5R = 30

85 - P - 5R = 30

85 - 30  = P + 5R

P + 5R = 55     (4)

From equation (2):

100 - 2P - 3Q - 4R = 30    --- (2)

100 - 2P - 3(5) - 4R = 30

100 - 15 - 30 = 2P +4R

55 = 2P + 4R

2P + 4R = 55     (5)

From equation (4) and (5); we have:

P + 5R = 55     (4)

2P + 4R = 55     (5)

From equation (4);

Let P = 55 - 5R

Then, replace P = 55 - 5R in equation (5)

2(55 - 5R) + 4R = 55  

110 - 10R + 4R = 55

110 - 6R = 55

6R = 110 - 55

6R = 55

R = 55/6

R = 9.17

Substitute, the value of R in equation (5); we have:

2P + 4R = 55

2P + 4(9.17) = 55

2P  = 55 - 4(9.17)

2P  = 55 - 36.68

2P  =  18.32

P = 18.32/2

P = 9.16

Therefore, the outlet stream rates are as follows:

                  Mass Feed Rate   ( mol/hr)               Molar ratio

CO                  30                                                    0.281

H₂                    30                                                   0.281              

CH₃OH            9.16                                                0.085

C₂H₅OH            5                                                   0.047

C₃H₇OH           9.17                                                0.086

CO₂                 23.34                                              0.219

Total:                106.67

1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation:

am i too late?

explanation: