Question

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Answer 1

Answer:

a) percent yield of  CH₄  = 85.5 %

b) percent yield of  AgBr = 79 %

c) percent yield of  FeCl₃ =  82 %

Explanation:

Part a:

Data given:

mass of carbon (C) = 1.4 g

actual yield of methane (CH₄) = 1.6 g

percent yield of methane (CH₄) = ?

Reaction Given:

                       C + 2H₂ -----> CH₄

Solution:

First we have to find theoretical yield.

So,

Look at the reaction

                          C + 2H₂ -----> CH₄

                      1 mol                   1 mol

As 1 mole of C  give 1 mole of CH₄

Convert moles to mass

molar mass of C = 12 = 12 g/mol

molar mass of  CH₄ = 12 + 4(1) = 16 g/mol

Now

                          C        +        2H₂     ------->     CH₄

                   1 mol (12 g/mol)                     1 mol (16 g/mol)

                          12 g                                       16 g

12 g of carbon (C) produce 16g of CH₄

So

if 12 g of carbon (C) produce 16g of CH₄ so how many grams of CH₄ will be produced by 1.4 g of C.

Apply Unity Formula

                           12 g of carbon (C) ≅ 16g of CH₄

                            1.4 g of carbon (C) ≅  X g of CH₄

Do cross multiply

                        mass of CH₄ = 16 g x 1.4 g / 20 g

                        mass of CH₄ = 1.87 g

So the Theoretical yield of  CH₄ = 1.87 g

Now Find the percent yield of CH₄

Formula Used

               percent yield = actual yield /theoretical yield x 100 %

Put value in the above formula

                        percent yield = 1.6 g / 1.87 g x 100 %

                         percent yield = 85.5 %

percent yield of  CH₄  = 85.5 %

______________________________

Part b:

Data given:

mass of Br₂ = 3.8 g

actual yield of AgBr = 6.7 g

percent yield of AgBr = ?

Reaction Given:

                       2Ag + Br₂ -----> 2AgBr

Solution:

First we have to find theoretical yield.

So,

Look at the reaction

                          2Ag + Br₂ -----> 2AgBr

                                     1 mol       2 mol

As 1 mole of Br₂ give 2 mole of AgBr

Convert moles to mass

molar mass of Br₂ = 2(80) = 160 g/mol

molar mass of  AgBr = 108 + 80 = 188 g/mol

Now

                          2Ag         +            Br₂       --------->      2AgBr

                                        1 mol (160 g/mol)           2 mol (188 g/mol)

                                                  160 g                            376 g

169 g of Br₂ produce 376 g of AgBr

So

if 169 g of Br₂ produce 376 g of AgBr then how many grams of  AgBr will be produced by 3.8 g of Br₂.

Apply Unity Formula

                           169 g of Br₂ ≅  376 g of AgBr

                           3.8 g of Br₂ ≅  X g of AgBr

Do cross multiply

                        mass of AgBr = 376 g x 3.8 g / 169 g

                        mass of AgBr = 8.45 g

So the Theoretical yield of AgBr = 8.45 g

Now Find the percent yield of AgBr

Formula Used

               percent yield = actual yield /theoretical yield x 100 %

Put value in the above formula

                        percent yield = 6.7 g / 8.45 g x 100 %

                         percent yield = 79 %

percent yield of  AgBr = 79 %

______________________________

Part c:

Data given:

mass of Cl₂ = 13 g

actual yield of FeCl₃ = 16.3 g

percent yield of FeCl₃ = ?

Reaction Given:

                       2Fe + 3Cl₂ -----> 2FeCl₃

Solution:

First we have to find theoretical yield.

So,

Look at the reaction

                           2Fe + 3Cl₂ -----> 2FeCl₃

                                     3 mol       2 mol

As 3 mole of Cl₂ give 2 mole of FeCl₃

Convert moles to mass

molar mass of Cl₂ = 2(35.5) = 71 g/mol

molar mass of  FeCl₃ = 56 + 3(35.5) = 162.5 g/mol

Now

                          2Fe         +            3Cl₂       --------->      2FeCl₃

                                        3 mol (71 g/mol)              2 mol (162.5 g/mol)

                                                  213 g                            325 g

213 g of Cl₂ produce 325 g of FeCl₃

So

213 g of Cl₂ produce 325 g of FeCl₃ then how many grams of FeCl₃ will be produced by 13 g of Cl₂.

Apply Unity Formula

                         213 g of Cl₂ ≅  325 g of FeCl₃

                        13 g of Cl₂ ≅ X g of FeCl₃

Do cross multiply

                        mass of FeCl₃ = 325 g x 13 g / 213 g

                        mass of FeCl₃ = 19.8 g

So the Theoretical yield of FeCl₃= 19.8 g

Now Find the percent yield of FeCl₃

Formula Used

               percent yield = actual yield /theoretical yield x 100 %

Put value in the above formula

                        percent yield = 16.3 g / 19.8 g x 100 %

                         percent yield = 82 %

percent yield of  FeCl₃ =  82 %