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11111nata11111 [884]

3 years ago

9

18 An important environmental consideration is the appropriate disposal of cleaning solvents. An environmental waste treatment c

ompany analysed a cleaning solvent, J, and found it to contain the elements carbon, hydrogen, and chlorine only. The chemical composition of J was determined using different analytical chemistry techniques. Combustion reaction: Combustion of 1.30 g of J gave 0.872 g CO2 and 0.089 g H2O. Precipitation reaction with AgNO3(aq): 0.535 g of J gave 1.75 g AgCl precipitate. (a) Determine the percentage by mass of carbon and hydrogen in J, using the combustion data. (3) (b) Determine the percentage by mass of chlorine in J, using the precipitation data. (1) (c) The molar mass was determined to be 131.38 g mol–1. Deduce the molecular formula of J.

1 answer:

Katyanochek1 [597]3 years ago

5 0

Answer:

a) Percentage by mass of carbon: 18.3%

Percentage by mass of hydrogen: 0.77%

b) Percentage by mass of chlorine: 80.37%

c) Molecular formula: $C_{2} H Cl_{3}$

Explanation:

Firstly, the mass of carbon must be determined by using a conversion factor:

$0.872g CO _{2} *\frac{12g C}{44g CO_{2} } = 0.238g CO_{2}$

The same process is used to calculate the amount of hydrogen:

$0.089g H_{2}O*\frac{2g H}{18g H_{2}O } = 0.010g H$

The percentage by mass of carbon and hydrogen are calculated as follows:

%C $\frac{0.238g}{1.3g} *100%$ = 18.3%

%H $\frac{0.010g}{1.3g} *100%$ =0.77%

From the precipation data it is possible obtain the amount of chlorine present in the compound:

$1.75 AgCl*\frac{35.45g Cl}{143.45g AgCl}$ = 0.43g AgCl

Let's calculate the percentage by mass of chlorine:

%Cl= $\frac{0.43g}{0.535g} * 100%$ = 80.37%

Assuming that we have 100g of the compound, it is possible to determine the number of moles of each element in the compound:

$18.3g C*\frac{1mol C}{12g C} = 1.52mol C$

$0.77g H*\frac{1mol H}{1g H} = 0.77mol H$

$80.37gCl*\frac{1molCl}{35.45g Cl} = 2.27mol Cl$

Dividing each of the quantities above by the smallest (0.77mol), the subscripts in a tentative formula would be

$C=\frac{1.52}{0.77} = 1.97$ ≈ 2

$H = \frac{0.77}{0.77} = 1$

$Cl =\frac{2.27}{0.77}=2.94$ ≈3

The empirical formula for the compound is:

$C_{2} H Cl_{3}$

The mass of this empirical formula is:

mass of C + mass of H + mass of Cl= 24g +1+ 106.35 =131.35g

This mass matches with the molar mass, which means that the supscript in the molecular formula are the same of the empirical one.

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___AsCl3+____H2S-->___As2S3+___HCI

Alex17521 [72]

Answer:

2 AsCl₃ + 3 H₂S → As₂S₃ + 6 HCl

Explanation:

When we balance a chemical equation, what we are trying to do is to achieve the same number of atoms for each element on both sides of the arrow. On the right of the arrow is where we can find the products, while the reactants are found on the left of the arrow.

We usually balance O and H atoms last.

AsCl₃ + H₂S → As₂S₃ +HCl

<u>reactants</u>

As --- 1

Cl --- 3

H --- 2

S --- 1

<u>products</u>

As --- 2

Cl --- 1

H --- 1

S --- 3

2 AsCl₃ + H₂S → As₂S₃ +HCl

<u>reactants</u>

As --- 2

Cl --- 6

H --- 2

S --- 1

<u>products</u>

As --- 2

Cl --- 1

H --- 1

S --- 3

The number of As atoms is now balanced.

2 AsCl₃ + 3 H₂S → As₂S₃ +HCl

<u>reactants</u>

As --- 2

Cl --- 6

H --- 6

S --- 3

<u>products</u>

As --- 2

Cl --- 1

H --- 1

S --- 3

The number of S atoms is now equal on both sides.

2 AsCl₃ + 3 H₂S → As₂S₃ + 6 HCl

<u>reactants</u>

As --- 2

Cl --- 6

H --- 6

S --- 3

<u>products</u>

As --- 2

Cl --- 6

H --- 6

S --- 3

The equation is now balanced.

3 0

3 years ago

Discussion: (24 points)

lorasvet [3.4K]

What exactly is this asking

6 0

3 years ago

Which of the answer choices correctly describes a substance that breaks apart in water to produce an excess of hydroxide ions (O

OverLord2011 [107]

The answer is a base which brings a lot of OH-

3 0

4 years ago

Determine the total pressure of all gases (at STP) formed when 50.0 mL of TNT (C3H5(NO3)3, , molar mass = 227.10 g/mol) reacts a

bija089 [108]

Answer:

Total pressure is 1189 atm

Explanation:

This is the reaction:

4C₃H₅(NO₃)₃ → 6N₂ + O₂ + 12CO₂ + 10H₂O

As we have the volume of TNT, we must know the density to find out the mass and then, apply molar mass to calculate mole.

TNT density = 1.654 g/mL

Density = mass / volume

1.654 g/mL = TNT mass / 50mL

1.654 g/mL . 50mL = TNT mass → 82.7 g

Mass / Molar mass = Mol → 82.7 g / 227.1 g/m = 0.364 mole

Now, we can calculate all the mole for the formed gases.

4 mole of TNT produce 6 mole N₂ ___ 1 mol O₂ __ 12 mole dioxide __ 10 mole of water

0.364 mole of TNT will produce:

- (0.364 . 6) /4 = 0.546 mole of produced nitrogen

- (0.364 . 1) /4 = 0.091 mole of produced oxygen

- (0.364 . 12) /4 = 1.092 mole of produced dioxide

- (0.364 . 10) /4 = 0.91 mole of produced vapour of water.

Total mole = 0.564 + 0.091 + 1.092 + 0.91 = 2.657 mole

Let's apply the Ideal Gases Law to find the total pressure, at STP

In STP, pressure is 1 atm for 1 mole at 273K, in a volume of 22.4 mL

But we have a volume of 50mL, and we have 2.657 total mole

Don't forget to convert 50 mL to L, cause the units for R

50 mL = 0.050L

P . 0.050L = 2.657 mol . 0.082L.atm/mol.K . 273K

P = (2.657 mol . 0.082L.atm/mol.K . 273K) / 0.050L

P = 1189 atm

3 0

3 years ago

6. 7. A hyperbaric chamber has a volume of 200. L. (a) How many moles of oxygen are needed to fill the chamber at room temperatu

Otrada [13]

Answer:

a) 24.7 mol

b) 790 g

Explanation:

Step 1: Given data

Volume of the chamber (V): 200. L

Room temperature (T): 23 °C

Pressure of the gas (P): 3.00 atm

Step 2: Convert "T" to Kelvin

We will use the following expression.

K = °C + 273.15

K = 23°C + 273.15 = 296 K

Step 3: Calculate the moles (n) of oxygen

We will use the ideal gas equation.

P × V = n × R × T

n = P × V/R × T

n = 3.00 atm × 200. L/(0.0821 atm.L/mol.K) × 296 K = 24.7 mol

Step 4: Calculate the mass (m) corresponding to 24.7 moles of oxygen

The molar mass (M) of oxygen ga sis 32.00 g/mol. We will calculate the mass of oxygen using the following expression.

m = n × M

m = 24.7 mol × 32.00 g/mol = 790 g

6 0

3 years ago