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


Answer 1


Mass CO2 = 3.75 grams

Mass H2O = 2.30 grams

Mass SO2 = 2.73 grams


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

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Atmospheric pressure arises due to the force exerted by the air above the Earth. At higher altitudes, the mass of the air above the Earth is _____ than at sea level, and atmospheric pressure therefore _____ with altitude.

If you perfom 30 joules of work lifting a 20-N box from the floor to a shelf how high is the shelf ​


Answer: 1.5 m

Explanation: E=mgh=Fh

30 J = 20 N * h

h = 30/20 = 1.5 m


1.5 m


it just is its easy made A on test

Ascorbic acid (vitaminc.Does not contain a traditional carboxylic acid group, but it is, nevertheless, still fairly acidic (pka = 4.2). Identify the acidic proton and explain your choice using resonance structures, if necessary:


Let us see the structure of ascorbic acid

As shown there is no COOH group however the OH group can lose a proton and forms conjugate base

The conjugate base formed is stabilized due to resonance

More the stability of conjugate base more the strength of acid

Hence ascorbic acid behaves as an acid

A student is given a sample of a blue copper sulfate hydrate. He weighs the sample in a dry covered porcelain crucible and got a mass of 23.875 g for the crucible, lid, and sample. The mass of the empty crucible and lid was found earlier to be 22.652 g. He then heats the crucible to expel the water of hydration, keeping the crucible at red heat for 10 minutes with the lid slightly ajar. On colling, he finds the mass of crucible, lid, and contents to be 23.403 g. The sample was changed in the process to very light clue anhydrous CuSO4. How many moles of water are present per mole of CuSO4?



There are present 5,5668 moles of water per mole of CuSO₄.


The mass of CuSO₄ anhydrous is:

23,403g - 22,652g = 0,751g.

mass of crucible+lid+CuSO₄ - mass of crucible+lid

As molar mass of CuSO₄ is 159,609g/mol. The moles are:

0,751g ×(1mol)/(159,609g) = 4,7052x10⁻³ moles CuSO₄

Now, the mass of water present in the initial sample is:

23,875g - 0,751g - 22,652g = 0,472g.

mass of crucible+lid+CuSO₄hydrate - CuSO₄ - mass of crucible+lid

As molar mass of H₂O is 18,02g/mol. The moles are:

0,472g ×(1mol)/(18,02g) = 2,6193x10⁻² moles H₂O

The ratio of moles H₂O:CuSO₄ is:

2,6193x10⁻² moles H₂O / 4,7052x10⁻³ moles CuSO₄ = 5,5668

That means that you have 5,5668 moles of water per mole of CuSO₄.

I hope it helps!

If the heat of combustion of carbon monoxide (CO) is −283.0 kJmole, how many grams of carbon monoxide must combust in order to release 2,500.kJ of energy?



247.4 g


Let's consider the thermochemical equation for the combustion of carbon monoxide.

CO(g) + 0.5 O₂(g) ⇒ CO₂(g)    ΔH°c = -283.0 kJ/mol

The moles of carbon monoxide required to release 2500 kJ (-2500 kJ) are:

-2500 kJ × (1 mol CO/-283.0 kJ) = 8.834 mol CO

The molar mass of CO is 28.01 g/mol. The mass corresponding to 8.834 moles of CO is:

8.834 mol × 28.01 g/mol = 247.4 g

Isn't this false? For the industrial production of indigo carmine, a blue food colouring additive, a synthetic process with an E-factor of 17.4 produces less waste than a synthetic process with an E-factor of 3.0.

The answer I got was False, is this correct?


Answer: yes it is false


The statement is false. A synthetic process with a lower E-factor produces less waste than a process with a higher E-factor.

The E-factor is a measure of the waste generated during a manufacturing process. It is calculated by dividing the total mass of waste produced by the mass of the desired product. A lower E-factor indicates that less waste is generated per unit of product.

In this case, the synthetic process with an E-factor of 3.0 produces less waste than the process with an E-factor of 17.4. This means that the process with an E-factor of 3.0 is more efficient in terms of waste reduction.

Rank each of the following gases in order of increasing Urms assuming equivalent amounts and all gases are at the same temperature and pressure where 1 is the lowest Urms Gas 1: H2S Gas 2: He Gas 3: NF3 Gas 4: H20



NF3< H2S< H2O< He


The average speed of a gas depends on the relative molecular mass of the gas. Lighter gases have a greater average speed and move faster than heavier gases.

Hence, we need to arrange these gases listed in the question in order of decreasing molecular mass in order to obtain the order of increasing Urms assuming equivalent amounts and all gases are at the same temperature and pressure.

Hence; NF3< H2S< H2O< He

Final answer:

The rank in order of increasing Urms for the gases given, assuming the same amount, temperature, and pressure is as follows: NF₃, H₂S, H₂O, and He.


Urms, or root mean square speed, of a gas, depends on the molar mass of the gas and the temperature. It is determined by the formula Urms = sqrt(3kT/M), where k is Boltzmann's constant, T is temperature, and M is the molar mass. The greater the molar mass, the lower the Urms. Looking at the molar masses of the gases H₂S (34 g/mol), He (4 g/mol), NF₃(71 g/mol), and H₂O (18 g/mol), we can rank them in order of increasing Urms as NF₃ 1st), H₂S (2nd), H₂O (3rd), and He (4th).

Learn more about Urms of gases here: