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mihalych1998 [28]
3 years ago
6

A stover-to-ethanol plant produces 40,000 tonne/yr of ethanol and contains 8functionalunits:feedstockhandling,pretreatment,simul

taneoussaccharification/fermentation distillation, solid/syrup separation, wastewatertreatment, boilers, and turbogeneration/utilities (McAloon et al., 2000). Estimatethe FCI of the plant using the empirical correlation based on the work ofBridgwater and Mumford (1979)
Chemistry
1 answer:
Step2247 [10]3 years ago
8 0

Answer:

FCI=458000*8*40000^{0.3}\\FCI=880818398.52 M

FCI=88.0818 MM≅88 MM

Explanation:

Empirical correlation based on the work of Bridgwater and Mumford (1979):

For Liquid or solid phase Plants:

FCI=458000*N*F^{0.3}                       F<60,000 tonne/yr     Eq (1)

FCI=7000*N*F^{0.68}                         F≥60,000 tonnes/yr   Eq (2)

Where:

N is the number of functional units

F is the process throughput tonnes/yr

In our case F=40,000 tonne/yr <60,000 tonne/yr, We are going to use Eq (1)

FCI=458000*N*F^{0.3}                       F<60,000 tonne/yr

N=8, F=40,000 tonne/yr

FCI=458000*8*40000^{0.3}\\FCI=880818398.52 M

FCI=88.0818 MM≅88 MM

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7 0
3 years ago
How many times more hydroxide ions are there in a solution with a ph of 9 than in a solution with a ph of 3?
cestrela7 [59]
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4 0
3 years ago
a 20.5g sample of cleaning detergent contains 8.61g of NH40H.CALCULATE the percentage composition of nitrogen in the cleaning de
AysviL [449]

Answer:

The mass percentage composition of nitrogen in the sample of the cleaning detergent is approximately 16.78%  

Explanation:

The given mass of the sample of the cleaning detergent, m₁ = 20.5 g

The mass of the ammonium hydroxide, NH₄OH in the detergent, m₂ = 8.61 g

The molar mass of NH₄OH = 35.04 g/mol

The molar mass of nitrogen, N = 14.01 g/mol

Therefore, the mass, m₃ of nitrogen, N, in 8.61 g of ammonium hydroxide, NH₄OH, is found as follows;

m₃ = (14.01/35.04) × 8.61 g = (402,087/118,800) g ≈ 3.44 g

The mass of nitrogen, N, in the ammonium hydroxide, NH₄OH, contained in the 20.5 g sample of the cleaning agent, m₃ ≈ 3.44 grams

The percentage composition of nitrogen in the sample of the cleaning detergent, %N is given as follows;

\% Composition = \dfrac{Mass \ of \ component}{Total \ mass \ of \ cleaning \ detergent} \times 100

Therefore;

%N ≈ ((3.44 g)/(20.5 g)) × 100 ≈ 16.78 %

The percentage composition of nitrogen, %N ≈ 16.78%.

6 0
3 years ago
Ethylene oxide (EO) is prepared by the vapor-phase oxidation of ethylene. Its main uses are in the preparation of the antifreeze
Rashid [163]

Answer:

a. ΔH^0_{rxn} = -108.0\frac{kJ}{mol}

b. 320.76° C

Explanation:

a.)

we can solve this type of question (i.e calculate ΔH^0_{rxn} , for the gas-phase reaction )  using the Hess's Law.

ΔH^0_{rxn} =  E_{product} deltaH^0_{t}-E_{reactant} deltaH^0_{t}

Given from the question, the table below shows the corresponding  ΔH^0_{t}(kJ/mol) for each compound.

Compound                    H^0_{t}(kJ/mol)

Liquid EO                       -77.4

CH_4_(g_)                            -74.9                

CO_(g_)                              -110.5

If we incorporate our data into the above previous equation; we have:

ΔH^0_{rxn} = (-110.5 kJ/mol + (-74.9 kJ/mol) ) - (-77.4 kJ/mol)

          =   -108.0 \frac{kJ}{mol}

b.)

We are to find the final temperature if the average specific heat capacity of the products is 2.5 J/g°C

Given that:

the specific heat capacity (c) = 2.5 J/g°C

T_{initial} = 93.0°C   &

the  enthalpy of vaporization  (ΔH^0_{vap}) = 569.4 J/g

If, we recall; we will remember that; Specific Heat Capacity is the amount of heat needed to raise the temperature of one gram of a substance by one kelvin.

∴ the specific heat capacity (c) is given as =  \frac{Heat(q)}{mass*changeintemperature(T_{initial}-T_{final})}

Let's not forget as well, that  ΔH^0_{vap} = \frac{q}{mass}

If we substitute  ΔH^0_{vap}  for  \frac{q}{mass} in the above equation, we have;

specific heat capacity (c) = \frac{deltaH^0_{vap}}{T_{final}-T_{initial}}

Making (T_{final}- T_{initial}) the subject of the formula; we have:

T_{final}- T_{initial}  = \frac{delat H^0_{vap}}{specificheat capacity}

(T_{final}-93.0^0C)=\frac{569.4J/g}{2.5J/g^0C}

T_{final}=\frac{569.4J/g}{2.5J/g^0C}+93.0^0C

         = 227.76°C +93.0°C

          = 320.76°C

∴ we can thereby conclude that the final temperature = 320.76°C                

7 0
3 years ago
20.0 g of bromic acid, HBrO3, is reacted with excess HBr.
Blababa [14]

After Rounding off The percentage yield is 64%

<h3>What is Percentage Yield ?</h3>

It is the ratio of actual yield to theoretical yield multiplied by 100% .

It is given in the question

20.0 g of bromic acid, HBrO3, is reacted with excess HBr.

The reaction is

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After Rounding off The percentage yield is 64% .

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5 0
2 years ago
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