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 %
a is the answer because all of the other answers are wrtong
Answer:
hey just where are the rections keep a pic
Explanation:
okay?.........................................
Explanation:
Expression for rate of the given reaction is as follows.
Rate = k[HgCl_{2}]x [C_{2}O^{2-}_{4}]y[/tex]
Therefore, the reaction equations by putting the given values will be as follows.
............. (1)
........... (2)
............ (3)
Now, solving equations (1) and (2) we get the value of y = 2. Therefore, by solving equation (2) and (3) we get the value of x = 1.
Therefore, expression for rate of the reaction is as follows.
Rate = ![k[HgCl_{2}]x [C_{2}O^{2-}_{4}]y](https://tex.z-dn.net/?f=k%5BHgCl_%7B2%7D%5Dx%20%5BC_%7B2%7DO%5E%7B2-%7D_%7B4%7D%5Dy)
Rate = ![k [HgCl2]1 [C_{2}O^{-2}_{4}]2](https://tex.z-dn.net/?f=k%20%5BHgCl2%5D1%20%5BC_%7B2%7DO%5E%7B-2%7D_%7B4%7D%5D2)
Hence, total order = 1 + 2 = 3
According to equation (1),
k =
Thus, we can conclude that rate constant for the given reaction is
.