Consider this reaction, which occurs in the atmosphere and contributes to photochemical smog: CH4(g) + CCl4(g) →2CH2Cl2(g) If there is 11.45 g CH4 and excess CCl4 present, the reaction yields 94.1 g CH2Cl2. Calculate the percent yield for the reaction.

The percent yield of the reaction is 77.6 %

Explanation:

Step 1: Data given

Mass of CH4 = 11.45 grams

Molar mass of CH4 = 16.04 g/mol

Mass of CH2Cl2 produced = 94.1 grams

Step 2: The balanced equation

CH4(g) + CCl4(g) →2CH2Cl2(g)

Step 3: Calculate moles CH4

Moles CH4 = mass CH4 / molar mass CH4

Moles CH4 = 11.45 grams / 16.04 g/mol

Moles CH4 =  0.714 moles

Step 4: Calculate moles CH2Cl2

For 1 mol CH4 we need 1 mol CCl4 to produce

For 0.714 moles CH4 we'll have 2*0.714 = 1.428 moles CH2Cl2

Step 5: Calculate mass CH2Cl2

Mass CH2Cl2 = moles CH2Cl2 * molar mass CH2Cl2

Mass CH2Cl2 = 1.428 moles * 84.93 g/mol

Mass CH2Cl2 = 121.3 grams

Step 6: Calculate percent yield

% yield = (actual yield / theoretical yield ) * 100%

% yield = (94.1 grams / 121.3 grams ) * 100 %

% yield = 77.6 %

The percent yield of the reaction is 77.6 %

each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i  

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