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Anna11 [10]
3 years ago
11

Methanol (ch3oh), also called methyl alcohol, is the simplest alcohol. it is used as a fuel in race cars and is a potential repl

acement for gasoline. methanol can be manufactured by combination of gaseous carbon monoxide and hydrogen. suppose 68.5 kg co(g) is reacted with 8.60 kg h2(g). calculate the theoretical yield of methanol. if 3.57 × 104 g ch3oh is actually produced, what is the percent yield of methanol?
Chemistry
1 answer:
erica [24]3 years ago
5 0

Answer:

             %age Yield  =  51.45 %

Solution:

Step 1: Convert Kg into g

68.5 Kg CO  =  68500 g CO

8.60 Kg H₂  =  8600 g

Step 2: Find out Limiting reactant;

The Balance Chemical Equation is as follow;

                                 CO  +  2 H₂    →    CH₃OH

According to Equation,

                   28 g (1 mol) CO reacts with  =  4 g (2 mol) of H₂

So,

                    68500 g CO will react with  =  X g of H₂

Solving for X,

                    X  =  (68500 g × 4 g) ÷ 28 g

                    X  =  9785 g of H₂

It shows 9785 g H₂ is required to react with 68500 g of CO but we are provided with 8600 g of H₂ which is less than required. Therefore, H₂ is provided in less amount hence, it is a Limiting reagent and will control the yield of products.

Step 3: Calculate Theoretical Yield

According to equation,

            4 g (2 mol) H₂ reacts to produce  =  32 g (1 mol) Methanol

So,

                          8600 g H₂ will produce  =  X g of CH₃OH

Solving for X,

                    X  =  (8600 g × 32 g) ÷ 4 g

                     X =  68800 g of CH₃OH

Step 4: Calculate %age Yield

                     %age Yield  =  Actual Yield ÷ Theoretical Yield × 100

Putting Values,

                     %age Yield  =  3.54 × 10⁴ g ÷ 68800 g × 100

                     %age Yield  =  51.45 %


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Explanation:

a) The mass fractium of helium is obtained converting the moles of the four gases to grams with molar weight and then caculating of the total of grams how many are of helium, thus:

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<em />

<em>The mass fraction of Helium is 1,6%</em>

<em />

<em>b)</em><em>  </em>Because the mole fraction of all gases is the same the average molecular weight of the mixture is:

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c) The molar concentration is possible to know ussing ideal gas law, thus:

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T is temperature: 500 K

And M is molar concentration. Replacing:

M = 3,6×10⁻² M

d) The mass density is possible to know converting the moles of molarity to grams with average molecular weight and liters to mililiters, thus:

3,6×10⁻² \frac{mol}{L} × \frac{64,775 g}{mol} × \frac{1L}{1000 mL} =

2,3×10⁻³ g/mL

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