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shepuryov [24]
3 years ago
11

PLZZZZZ HELP MEEEEEE

Chemistry
1 answer:
Masja [62]3 years ago
4 0

Answer:

0.0745 mole of hydrogen gas

Explanation:

Given parameters:

Number of H₂SO₄ = 0.0745 moles

Number of moles of Li = 1.5107 moles

Unknown:

Number of moles of H₂ produced = ?

Solution:

To solve this problem, we have to work from the known specie to the unknown one.

The known specie in this expression is the sulfuric acid,  H₂SO₄. We can compare its number of moles with that of the unknown using a balanced chemical equation.

   Balanced chemical equation:

                    2Li   +     H₂SO₄   →   Li₂SO₄   +   H₂

 From the balanced equation;

     

Before proceeding, we need to obtain the limiting reagent. This is the reagent whose given proportion is in short supply. It determines the extent of the reaction.

           2 mole of Li reacted with 1 mole of  H₂SO₄

          1.5107 mole of lithium will react with \frac{1.5107}{2}  = 0.7554mole of H₂SO₄

But we were given 0.0745 moles,

This suggests that the limiting reagent is the sulfuric acid because it is in short supply;

   

   since 1 mole of sulfuric acid produced 1 mole of hydrogen gas;

    0.0745 mole of sulfuric acid will produce 0.0745 mole of hydrogen gas

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A single-effect evaporator is concentrating a feed of 9072 kg/h of a 10 wt % solution of NaOH in water at temperature of 288.8 K
Aloiza [94]

Answer:

a) steam used = 8440 kg/hr

b)  Because as per calculation done in a), Steam is sufficient to evaporate 7257.6 kg/hr water from feed. Since feed is less than this amount, therefore whole water will be evaporated and 100% solid product will be obtained in bottom collection

Explanation:

Saturated steam pressure = 42KPag=1.42 bar

From steam table, Steam temperature = 110 C

Latent heat of this steam = 2230 KJ/kg

Process side pressure = 20 Kpaa = .2 bar

Water latent heat at this Pressue = 2360 KJ/kg

Water boiling Point at this Pressure =60 C

Feed Inlet temperature = 15.6 C

Total Heat required = Heat required to rise feed temperature from 15.6 to 60 degree C + Evaporation of water to concentrate the feed

Total Water evaporation required = 9072*(100-10)/100-9072*.1/.5*.5=7257.6 Kg/hr

Specific heat of feed assumed = 4.2 KJ/kg/K

=> Total heat required = 9072*4.2*(60-15.6)+7257.6*2360=18819682 KJ/hr

LMTD = ((110-60)-(110-15.6))/LN((110-60)/(110-15.6))=69.8 C

U, Overall Heat transfer coefficient = 1988 W/m2/K

Total heat required = U*A*LMTD

=> Area required of evaporator, A=18819682 *1000/3600/1988/59.8=44 m2

Steam used = 18819682/2230=8440 kg/hr

Energy required to condense vaporised feed = 7257.6*2360=17127936 KJ/hr

=> Steam efficiency = (18819682-17127936)/18819682=9%

b) Because as per calculation done in a), Steam is sufficient to evaporate 7257.6 kg/hr water from feed. Since feed is less than this amount, therefore whole water will be evaporated and 100% solid product will be obtained in bottom collection

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4 years ago
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