Answer to Question #212327 in Chemistry for Tlholo

Question

Methanol(CH3OH)is the simplest alcohol. It is used as a fuel in race cars and Is a potential replacement for petrol .Methanol can be manufactured by combining gaseous carbon monoxide and hydrogen when 68.5kg COis reacted with 8.60kg H2 ,3.57×10⁴g CH3OH is actually produced.

Answer 1

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 = 51.45 %

Hence the percent yield of methanol is 51.45%