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2 years ago

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A reaction was performed in which 3.7 g of benzoic acid was reacted with excess methanol to make 2.1 g of methyl benzoate. Calcu

late the theoretical yield and percent yield for this reaction.

1 answer:

Sav [38]2 years ago

4 0

Answer:

The theoretical yield is 4.13 grams methyl benzoate

The percent yield is 50.8 %

Explanation:

Step 1: Data given

Mass of benzoic acid = 3.7 grams

Mass of methyl benzoate = 2.1 grams

Molar mass of benzoic acid = 122.12 g/mol

Molar mass of methyl benzoate = 136.15 g/mol

Step 2: The balanced equation

C7H6O2 + CH3OH → C8H8O2 + H2O

Step 3: Calculate moles benzoic acid

Moles benzoic acid = mass benzoic acid / molar mass

Moles benzoic acid = 3.7 grams / 122.12 g/mol

Moles benzoic acid = 0.0303 moles

Step 4: Calculate moles methy benzoate

For 1 mol benzoic acid we'll have 1 mol methyl benzoate

For 0.0303 moles benzoic acid we'll have 0.0303 moles methyl benzoate

Step 5: Calculate mass methyl benzoate

Mass methyl benzoate = moles methyl benzoate * molar mass

Mass methyl benzoate = 0.0303 moles * 136.15 g/mol

Mass methyl benzoate = 4.13 grams

Step 6: Calculate % yield

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

% yield = (2.1 grams / 4.13 grams ) *100%

% yield = 50.8 %

The theoretical yield is 4.13 grams methyl benzoate

The percent yield is 50.8 %

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Read 2 more answers

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1) Area of one side of each cube

The area of one side of each cube can be calculated by multiplying the length times the width:

$A=L\cdot W$

where

L = length

W = width

For cube A: $L=1 cm, W=1 cm$

So the area is: $A_A = (1)(1)=1 cm^2$

For cube B: $L=2 cm, W=2 cm$

So the area is: $A_B=(2)(2)=4 cm^2$

For cube C: $L=3 cm, W=3 cm$

So the area is: $A_C=(3)(3)=9 cm^2$

2) Surface area of each cube

The surface area of a cube is given by the sum of the areas of all its faces. Since a cube has 6 identical faces, this means that the total surface area is equal to 6 times the area of one face:

$A=6 A_1$

where

$A_1$ is the area of one face of the cube

For cube A, the area of one face is $A_A=1 cm^2$

So the surface area of cube A is

$A'_A = 6A_A=6(1)=6 cm^2$

For cube B, the area of one face is $A_B=4 cm^2$

So the surface area of cube B is

$A'_B = 6A_B=6(4)=24 cm^2$

For cube C, the area of one face is $A_C=9 cm^2$

So the surface area of cube C is

$A'_C = 6A_C=6(9)=54 cm^2$

3) Volume of each cube

The volume of a cube is obtained by multiplying its length, its width and its height:

$V=L\cdot W \cdot H$

Where:

L = length

W = width

H = height

Moreover for a cube, all the sides have equal length, so $L=W=H$

So the volume can be rewritten as

$V=L^3$

For cube A: L = 1 cm

So the volume is $V_A=(1 cm)^3 = 1 cm^3$

For cube B: L = 2 cm

So the volume is $V_B=(2 cm)^3 = 8 cm^3$

For cube C: L = 3 cm

So the volume is $V_C = (3 cm)^3 = 27 cm^3$

4) Ratio surface area/volume for each cube

In this part, we have to calculate the ratio between surface area and volume of each cube:

$r=\frac{A}{V}$

where

A is the surface area

V is the volume

For cube A, we have:

$A_A = 6 cm ^2$ (surface area)

$V_A=1 cm^3$ (volume)

So the ratio for cube A is:

$r=\frac{6}{1}=6$

For cube B, we have:

$A_B = 24 cm ^2$ (surface area)

$V_B=8 cm^3$ (volume)

So the ratio for cube B is:

$r=\frac{24}{8}=3$

For cube C, we have:

$A_C = 54 cm ^2$ (surface area)

$V_C=27 cm^3$ (volume)

So the ratio for cube C is:

$r=\frac{54}{27}=2$

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