Liquid ethyl mercaptan, C2H6S, has a density of 0.84 g/mL. Assuming that the combustion of this compound produces only CO2 , H2O, and SO2 , what masses of each of these three products would be produced in the combustion of 3.15 mL of ethyl mercaptan

Answers

Answer 1
Answer:

Answer:

Mass CO2 = 3.75 grams

Mass H2O = 2.30 grams

Mass SO2 = 2.73 grams

Explanation:

Step 1: Data given

Density of Liquid ethyl mercaptan, C2H6S = 0.84 g/mL

Volume of ethyl mercaptan = 3.15 mL

Step 2: The reaction

2C2H6S + 9O2 → 4CO2 + 6H2O + 2SO2

Step 3: Calculate mass of ethyl mercaptan

Mass = Volume * density

Mass ethyl mercaptan = 3.15 mL * 0.84 g/mL

Mass ethyl mercaptan = 2.646 grams

Step 4: Calculate moles ethyl mercaptan

Moles = mass / molar mass

Moles ethyl mercaptan = 2.646 grams / 62.13 g/mol

Moles ethyl mercaptan = 0.04259 moles

Step 5: Calculate moles of other products

For 2 moles ethyl mercaptan we need 9 moles O2 to produce 4 moles CO2, 6 moles H2O and 2 moles SO2

For 0.04259 moles we need 0.1917 moles O2 to produce:

2*0.04259 = 0.08518 moles CO2

3*0.04259 = 0.1278 moles H2O

1*0.04259 = 0.04259 moles SO2

Step 6: Calculate mass produced

Mass = moles * molar mass

Mass CO2 = 0.08518 moles * 44.01 g/mol

Mass CO2 = 3.75 grams

Mass H2O = 0.1278 moles * 18.02 g/mol

Mass H2O = 2.30 grams

Mass SO2 = 0.04259 moles * 64.07 g/mol

Mass SO2 = 2.73 grams


Related Questions

Sodium tert-butoxide (NaOC(CH3)3) is classified as bulky and acts as Bronsted Lowry base in the reaction. It is reacted with 2-chlorobutane. Based on this information, the major organic product(s) of this reaction are expected to be _____.
A 135 g sample of H20 at 85°C is cooled. The water loses a total of 15 kJ of energy in the coolingprocess. What is the final temperature of the water? The specific heat of water is 4.184 J/g.°C.A. 112°CB. 58°CC. 70°CD. 84°CE. 27°C
The Heat of vaporization for NH3= 1360 J/g. Calculate the quantity of heat energy (in kJ)needed to completely boil a 155 gram sample at its boiling pointgeneral formula
When the following redox equation is balanced with smallest whole number coefficients, the coefficient for Sn(OH)3– will be _____. Bi(OH)3(s) + Sn(OH)3–(aq) → Sn(OH)62–(aq) + Bi(s) (basic solution)
Estimate how much heat in joules is released when 25.0 g of water (C = 4.184 J/g°C) is cooled from 80.0°C to 30.0°C?

How many lone pairs are on the central atom of BrF3?

Answers

According to the molecular geometry, there are two lone pairs on central atom of BrF₃.

What is molecular geometry?

Molecular geometry can be defined as a three -dimensional arrangement of atoms which constitute the molecule.It includes parameters like bond length,bond angle and torsional angles.

It influences many properties of molecules like reactivity,polarity color,magnetism .The molecular geometry can be determined by various spectroscopic methods and diffraction methods , some of which are infrared,microwave and Raman spectroscopy.

They provide information about geometry by taking into considerations the vibrational and rotational absorbance of a substance.Neutron and electron diffraction techniques provide information about the distance between nuclei and electron density.

Learn more about molecular geometry,here:

brainly.com/question/28557524

#SPJ2

Answer:

BrF3 has 2 lone pairs (4 unshared electrons)

Explanation:

4Ga + 3S2 → 2Ga2S3 1. How many grams of Gallium Sulfide would form if 20.5 moles of Gallium burned?

Answers

Answer:

m_(Ga_2S_3)=2415.31gGa_2S_3

Explanation:

Hello,

In this case, for the given chemical reaction, we can notice there is a 4:2 molar ratio between the burned moles of gallium and the yielded moles of gallium sulfide, therefore, we compute them as shown below:

n_(Ga_2S_3)=20.5molGa*(2molGa_2S_3)/(4molGa)=10.25mol Ga_2S_3

Then, by using the molar mass of gallium sulfide (235.64 g/mol), we directly compute the grams:

m_(Ga_2S_3)=10.25mol Ga_2S_3*(235.64gGa_2S_3)/(1molGa_2S_3) \n\nm_(Ga_2S_3)=2415.31gGa_2S_3

Best regards.

What type(s) of intermolecular forces exist between two molecules of 1-pentanol? (BLB Ch. 11)

Answers

Answer:

Hydrogen bonding

London dispersion forces

Explanation:

The most pronounced and consequential intermolecular forces at play has to be hydrogen bonding due to the dipole created between the oxygen and hydrogen in the hydroxyl group. And then to a lesser extent there are very weak London dispersion forces that are always going to be there in between any two molecules of any species.

Many drugs are sold as their hydrochloric salts (r2nh2+cl−), formed by reaction of an amine (r2nh) with hcl. part 1 out of 4 draw the major organic product formed from the formation of acebutolol with hcl. acebutolol is a β blocker used to treat high blood pressure. omit any inorganic counterions.

Answers

Answer: -

Many drugs are sold as their hydrochloric salts (RNH₃⁺Cl⁻), formed by reaction of an amine (RNH₂) with HCl.

It is done because amines are generally liquids. But their hydrochloric salts are solid. A solid drug is always more preferable for drug companies as their handling and packaging are easier.

Acebutolol consists of one amide functionality as well as a secondary amine functionality.

When HCl is added, the lone pairs of the nitrogen of the secondary amine attacks it, leading to the formation of it's hydrochloric salts.

A gas occupies a volume of 30.0L, a temperature of 25°C and a pressure of 0.600atm. What will be the volume of the gas at STP?​

Answers

Answer:

=16.49 L

Explanation:

Using the equation

P1= 0.6atm V1= 30L, T1= 25+273= 298K, P2= 1atm, V2=? T2= 273

P1V1/T1= P2V2/T2

0.6×30/298= 1×V2/273

V2=16.49L

Which is more water soluble hexanoic acid or sodium hexanoate?

Answers

Answer:

Explanation:

Hexanoic acid [CH 3(CH 2) 4COOH] is barely soluble in water (about 1.0 g/100 g of water). Palmitic acid [CH 3(CH 2) 14COOH], with its large nonpolar hydrocarbon component, is essentially insoluble in water. The carboxylic acids generally are soluble in such organic solvents as ethanol, toluene, and diethyl ether.

hope this help ya~