Answer: 33.3 moles
Explanation: 67.2 g H2 = 67.2/2.016 = 33.3 moles
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 %
CH3 is the empirical formula for the compound.
A sample of a compound is determined to have 1.17g of Carbon and 0.287 g of hydrogen.
The number of atom or moles in the compound is
1.17 g C X 1 mol of C / 12.011 g C = 0.097411 mol of C.
0.287 g H x 1 mol of H / 1 g H = 0.28474 mol H.
This compound contains 0.097411 mol of carbon and 0.28474 mol of Hydrogen.
So we can represent the compound with the formula C0.974H0.284.
Subscripts in formulas can be made into whole numbers by multiplying the smaller subscript by the larger subscript.
we can divide 0.284 by 0.0974.
0.284 / 0.0974 = 3.
So here, Carbon is one and hydrogen is 3.
We can write the above formula as a CH3.
Hence the empirical formula for the sample compound is CH3.
For a detailed study of the empirical formula refer given link brainly.com/question/13058832.
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