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2 years ago

What is the theoretical yield of C6H5Br if 42.1 g of C6H6 react with 73.0 g of Br2?

1 answer:

3 0

1) Chemical reaction

C6H6 + Br2 ---> C6H5Br + HBr

2) Molar raios

1 mol C6H6 : 1 mol Br2 : 1 mol C6H5Br

3) Convert the data to moles

42.1 g of C6H6

molar mass of C6H6 = 6*12g/mol + 6*1g/mol = 78 g/mol

42.1 g / 78 g/mol = 0.54 mol

73.0 g of Br2

molar mass of Br2 = 2 * 79.9 g/mol = 159.8 g/mol

73.0 g / 159.8 g/mol = 0.46 mol of Br2

=> limiting reagent is the Br2.

4) Product

1 mol of Br2 (limiting reagent) yields 1 mole of C6H5Br, then you will obtain 0.46 mol of C6H5Br

5) Convert the product to grams

molar mass of C6H5Br = 6*12g/mol + 5*1g/mol + 79.9 g/mol = 156.9 g/mol

0.46mol*156.9g/mol = 72.2 g.

Answer: 72.2 g

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A large cylindrical tank contains 0.750 m3 of nitrogen gas at 27°C and 7.50 * 103 Pa (absolute pressure). The tank has a tight‐f

Nuetrik [128]

Answer: The answer is 68142.4 Pa

Explanation:

Given that the initial properties of the cylindrical tank are :

Volume V1= 0.750m3

Temperature T1= 27C

Pressure P1 =7.5*10^3 Pa= 7500Pa

Final properties of the tank after decrease in volume and increase in temperature :

Volume V2 =0.480m3

Temperature T2 = 157C

Pressure P2 =?

Applying the gas law equation (Charles and Boyle's laws combined)

P1V1/T1 = P2V2/T2

(7500 * 0.750)/27 =( P2 * 0.480)/157

P2 =(7500 * 0.750* 157) / (0.480 *27)

P2 = 883125/12.96

P2 = 68142.4Pa

Therefore the pressure of the cylindrical tank after decrease in volume and increase in temperature is 68142.4Pa

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2 years ago

Are The Reactions Of Ordinary Molecular Hydrogen Slow Or Rapid ? Why ?

olchik [2.2K]

In nature reactions of ordinary molecular hydrogen are slow since it's a diatomic molecule whose atoms are held together by very strong covalent bonds.The reaction rate of hydrogen varies depending on temperature and the properties of the reactants, for instance under high temperatures above 500°C hydrogen reacts vigorously and with fluorine it reacts explosively even under low temperatures

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2 years ago

How many chloride ions are in 0.486 moles of chloride ions?

svetlana [45]

Answer:

Since in a chloride ion, we have an additional electron

you might think that it will affect the mass but the mass of an electron is almost negligible so we will ignore that

Amount of ions in 1 mol = 6.022 * 10^23

Amount of ions in 0.486 moles = 0.486 * (6.022*10^23)

Amunt of ions in 0.486 moles = 2.9 * 10^23 ions

Hence, option 1 is correct

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2 years ago

ethylene glycol used in automobile antifreeze and in the production of polyester. The name glycol stems from the sweet taste of

Luden [163]

Answer: The empirical and molecular formula for the given organic compound is $CH_3O$ and $C_4H_{12}O_4$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=9.06g$

Mass of $H_2O=5.58g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 9.06 g of carbon dioxide, $\frac{12}{44}\times 9.06=2.47g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 5.58 g of water, $\frac{2}{18}\times 5.58=0.62g$ of hydrogen will be contained.

Mass of oxygen in the compound = (6.38) - (2.47 + 0.62) = 3.29 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{2.47g}{12g/mole}=0.206moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.62g}{1g/mole}=0.62moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{3.29g}{16g/mole}=0.206moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.206 moles.

For Carbon = $\frac{0.206}{0.206}=1$

For Hydrogen = $\frac{0.62}{0.206}=3$

For Oxygen = $\frac{0.206}{0.206}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 3 : 1

Hence, the empirical formula for the given compound is $C_1H_{3}O_1=CH_3O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 124 amu = 124 g/mol

Mass of empirical formula = 31 g/mol

Putting values in above equation, we get:

$n=\frac{124g/mol}{31g/mol}=4$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 4)}H_{(3\times 4)}O_{(1\times 4)}=C_4H_{12}O_4$

Thus, the empirical and molecular formula for the given organic compound is $CH_3O$ and $C_4H_{12}O_4$

3 0

2 years ago

A curium-245 nucleus is hit with a neutron and changes as shown by the equation. Complete the equation by filling in the missing

Lera25 [3.4K]

Try this option:

$^{245}_{96}Cm+^1_0n=^{103}_{42}Mo+^{245+1-103-3*1}_{96+0-42-3*0}(X)+3^1_0n;$

$^{245}_{96}Cm+^1_0n=> ^{103}_{42}Mo+^{140}_{54}(X)+3^1_0n;$

$^{245}_{96}Cm+^1_0n=> ^{103}_{42}Mo+^{140}_{54}Xe+3^1_0n;$

Answer:

$^{140}_{54}Xe$

8 0

2 years ago

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