From diagram, assuming that the reaction is 40% complete in the vessel: a. The weight of the methanol added per day b. The amount of Hl recycled Here is my solution on the problem and the diagram below is the actual given of the problem. I hope you could help me correct it.

• What's your question? May 13 '20 at 19:36
• (1) I assume that this is a check my work question? (2) Were you given any information other than what is in the diagram?
– MaxW
May 13 '20 at 22:21
• Note the posting a picture of written text instead of typing that text is discouraged for several reasons. 1/It cannot be searched for. 2/It cannot be reused as reference or quoting, putting extra effort on shoulders of responders. 3/It can be challenging to read and interpret. 4/ Low effort of question preparation leads often to low effort of question answering, even to question closure. May 14 '20 at 5:43

Edit 5/16 3pm Another equation just occurred to me...

Edit 5/16 6pm fixed a math mistake in equation (2)

This isn't an answer yet, I'm trying to clarify the question.

(1) I assume that this is a check my work question?

(2) Were you given any information other than what is in the diagram below? So using the variables indicated in the diagram:

R = recycled HI, units = lb/day
X = feed of $$\ce{CH3OH}$$, units = lb/day
W = waste stream, units = lb/day
Y = product stream, units = lb/day

• The waste and product outputs give % values. Weight % or Mole fraction %? Weight % would seem to be the right choice to me.

• The diagram would seem to indicate that the total input to the reactor is $$\pu{2000 + R lbs/day}$$ of $$\ce{HI}$$.

(A) - 2000 lb/day of HI feed

Assume 2000 lb/day of $$\ce{HI}$$ is input into the reactor plus whatever is recycled. We can only assume that the recycling has reached a steady state, so 2000 lbs/day of HI, or its equivalent, must be in the input stream to the separator. So summing the two steams from the separator:

\begin{align} \pu{1 lb \ce{HI}} &= \pu{1 lb \ce{HI}} \times \pu{2200 g/lb} \times \dfrac{\pu{141.939 g \ce{CH3I}/mol}}{\pu{127.911 g \ce{HI}/mol}}\times \dfrac{1}{\pu{2200 g/lb}} \\ &= \pu{1.1097 lb \ce{CH3I}} \\ \end{align}

$$\therefore 2000 = 0.8265\times W + \dfrac{0.816}{1.1097}\times Y = 0.8265\times W + 0.73533\times Y\tag{1}$$

(B) Now for $$\ce{CH3OH}$$ feed

$$\pu{1 lb \ce{CH3OH}} = \dfrac{141.939}{32.04} = \pu{4.4301 lb \ce{CH3I}}$$

$$\therefore X = \dfrac{0.816}{4.4301}\times Y + 0.184 \times Y = 0.36819 \times Y\tag{2}$$

(C) Balance water and $$\ce{CH3I}$$

The moles of water in the waste stream has to equal the moles of $$\ce{CH3I}$$ in the product stream.

$$\dfrac{0.1735\times W}{18.015} = \dfrac{0.816\times Y}{141.94}$$

$$\therefore W = 0.59693 \times Y\tag{3}$$

Substituting (3) into (1)

$$2000 = 0.8265\times (0.59693 \times Y) + 0.73533\times Y\tag{1a}$$

or $$Y = 1,627.75 \approx \pu{1628 lbs/day}$$
and $$W = 0.59693\times 1627.75 = 971.65 \approx \pu{972 lbs/day}$$
and $$X = 0.36819 \times 1,627.75 = 599.32 \approx \pu{599 lbs/day}$$

(D) For R -- "40% Complete" of what?

• 40% of HI is reacted?
• 40% of $$\ce{CH3OH}$$ is converted to $$\ce{CH3I}$$?

If the 40% percent is for the fraction of $$\ce{CH3OH}$$ which reacted, then 81.6:18.4 is not 60:40 in weight % or mole per cent for $$\ce{CH3I:CH3OH}$$.

$$\begin{array}{|c|c|c|c|c|}\hline & \pu{M.W.} &\pu{mol \%} & \pu{\dfrac{\pu{g}}{\pu{mol}}} & \pu{wt \%} \\ \hline \ce{CH3I} & 141.939 & 82.65 & 117.31 & 95.47 \\ \hline \ce{CH3OH} & 32.04 & 17.35 & 5.56 & 4.53 \\ \hline \end{array}$$

$$\begin{array}{|c|c|c|c|c|}\hline & \pu{M.W.} &\pu{wt \%} & \pu{\dfrac{\pu{mol}}{\pu{100 g}}} & \pu{mol \%} \\ \hline \ce{CH3I} & 141.939 & 82.65 & 0.58229 & 51.814 \\ \hline \ce{CH3OH} & 32.04 & 17.35 & 0.54151 & 48.186 \\ \hline \end{array}$$

The only way to use the information seems to be to assume that 40% of the total $$\ce{HI}$$ feed, 2000 + R, is converted to $$\ce{CH3I}$$.

$$\therefore (2000 +R)\times 0.4 = \dfrac{0.816}{1.1097}\times Y = 0.73533\times Y\tag{4}$$

Using Y from above, $$R = 992.33 \approx \pu{992 lbs/day}$$

Note - In comments below user Chet Miller makes a convincing argument about two points.

• The 40% reacted would refer to the % of the limiting reactant which reacted. This would mean that the 40% referred to $$\ce{CH3OH}$$.

• The OP's diagram is incorrectly drawn. Being a chemical engineer he proposes a diagram which makes more sense. There has to be a second separator of some sort.

I doubt that you could get all the HI out of water stream, but concentrated HI is about 57% HI by weight. So the reactor output and the waste stream, W, would have to be above atmospheric pressure.

Given Chet's proposed diagram below $$R = 0.8265\times W = 803.06 \approx \pu{803 lbs/day}$$ • Shouldn't the recycle stream be coming from the separator, and have the same concentration as in the waste stream? For 40% complete, they mean that the limiting reactant is 40% consumed. May 15 '20 at 12:16
• @ChetMiller - RE: "recycle stream be coming from the separator" ---I don't know. I took the diagram literately, without making any assumptions about the actual process. Obviously some kind of separation would have to be happening if just HI was recycled from the reactor output. That being said I don't think you'd recycle waste from the separator shown in the diagram. You wouldn't want to put the water back into the reactor.
– MaxW
May 15 '20 at 17:10
• RE: 40% complete, they mean that the limiting reactant is 40% consumed. --- Could be, I'm not extremely familiar with how chem engineering problems are setup. I also wonder how accurate the OP's drawing is.
– MaxW
May 15 '20 at 17:13

It's much easier to work in terms of lb-mole flow rates than in terms of lb or in terms of metric units. So I will express X,Y, W, and R in terms of lb-moles/day. 2000 lb/day of HI is 15.6 lb-moles/day. Since the stoichiometry is such the number of moles of product from the reaction are equal to the number of moles of reactant, the molar flow must be constant. Therefore we can use molar balances on the system. The overall molar balance on the system is $$15.6+X=Y+W$$ In addition, a carbon balance on the system is $$X=Y$$ and thus $$W = 15.6\ lb-moles/day$$.

The mole fractions in stream Y are:$$CH_3I=0.4998$$ and $$CH_3OH=0.5002$$

The mole fractions in stream W are: $$HI=0.4012$$and$$H_2O=0.5998$$

So an overall balance on iodine yields: $$15.6=0.4012W+0.4988Y=(0.4012)(15.6)+0.4998W$$From this, it follows that $$Y=18.76\ lb-moles/day=X$$

These values agree with the mass flow rates determined by @MaxW.