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miv72 [106K]
3 years ago
7

In the industrial synthesis of hydrogen, mixtures of CO and H2 are enriched in H2 by allowing the CO to react with steam. The ch

emical equation for this so-called water-gas shift reaction is
Chemistry
1 answer:
saw5 [17]3 years ago
8 0

Answer:

CO(g) + H2(g) + H2O(g) ==> CO2(g) + 2H2(g)

Explanation:

In the industry, hydrogen is prepared from water and hydrocarbons. Water gas being the major method of preparation of hydrogen industrially.

The water-gas reaction is an industrial process in which steam is passed over red-hot coke giving a gaseous mixture of carbon monoxide and hydrogen:

C + H2O(g) → CO + H2.

The mixture of CO and H2 is Futher passed through steam according to the equation:

CO(g) + H2(g) + H2O(g) ==> CO2(g) + 2H2(g) to give hydrogen and carbon dioxide.

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The average propane cylinder for a residential grill holds approximately 18 kg of propane. how much energy (in kj) is released b
Angelina_Jolie [31]
Let's begin with the basic values  that will be used in the solution.

The formula of propane is C3H8. It is an alkane, a hydrocarbon with the general formula of CnH2n+2. Notice that hydrocarbons have only Carbon and Hydrogen atoms. Its molar mass (M) is 44 g.

Molar Mass Calculation is done as like that
C=12 g/mol, H=1 g/mol. 1 mole propane has 3 moles Carbon atoms and 8 mole Hydrogen atoms. M(
C3H8)= 3*12+ 8*1= 44 g

Combustion reaction of hydrocarbons gives carbon dioxide and water by releasing energy. That energy is called as enthalpy of combustion (
ΔHc°).  

ΔHc° of propane equals -2202.0 kj/mol. Burning of 1 mole C3H8 releases 2202 kj energy. Minus sign only indicates that the energy is given out ( an exothermic reaction ).

Let's write the combustion reaction.
C3H8 + O2 ---> CO2 + H20 (unbalanced) 
ΔHc° = -2202 kj/mol

Now, we calculate mole of 20 kg propane. Convert kilogram into gram since we use molar mass is defined in grams.
mole=mass/molar mass ; n=m/M ; n= 20000 g /44 (g/mol)=454 mole

1 mole propane releases 2202 kj energy.
454 mole propane release 2202 kj *454= 1000909 kj

The answer is 1000909 kj.



6 0
3 years ago
A mixture initially contains AA, BB, and CC in the following concentrations: [A][A]A_1 = 0.550 MM , [B][B]B_1 = 1.40 MM , and [C
Alex787 [66]

Answer:

The value of the equilibrium constant KC is 1.244

Explanation:

A mixture initially contains A, B, and C in the following concentrations: [A] = 0.550 M, [B] = 1.40 M, and [C] = 0.600 M. The following reaction occurs and equilibrium is established: A+2B<->C

At equilibrium, [A] = 0.430 M and [C] = 0.720 M. Calculate the value of the equilibrium constant, Kc

Step 1: The balanced equation

A+2B<->C

Step 2: The initial concentrations

[A] = 0.550 M

[B]= 1.40 M

[C] = 0.600 M

Step 3: The concentraions at equilibrium

[A] = 0.550 -X = 0.430 M

[B]= 1.40 -2X M

[C] = 0.600 + X = 0.720 M

X = 0.120 M

[A] = 0.550 - 0.120 = 0.430 M

[B]= 1.40 -2*0.120 =  1.16 M

[C] = 0.600 + 0.120 = 0.720 M

Step 4: Calculate Kc

Kc = [C] / [A][B]²

Kc = 0.720 / (0.430*1.16²)

Kc = 1.244

The value of the equilibrium constant KC is 1.244

5 0
3 years ago
Elemental sulfur is one of the products of the gas-phase reaction of nitric acid and hydrogen sulfide. The other
Goshia [24]

Answer:

The manufacturing processes for liquefied petroleum gas are designed so that the majority, if not all, of the sulfur compounds are removed. The total sulfur level is therefore considerably lower than for other crude oil-based fuels and a maximum limit for sulfur content helps to define the product more completely. The sulfur compounds that are mainly responsible for corrosion are hydrogen sulfide, carbonyl sulfide and, sometimes, elemental sulfur. Hydrogen sulfide and mercaptans have distinctive unpleasant odors. A control of the total sulfur content, hydrogen sulfide and mercaptans ensures that the product is not corrosive or nauseating. Stipulating a satisfactory copper strip test further ensures the control of the corrosion.

8 0
2 years ago
The molecular weight of a gas is ________ g/mol if 3.5 g of the gas occupies 2.1 l at stp
bija089 [108]
<span>Pre-1982 definition of STP: 37 g/mol Post-1982 definition of STP: 38 g/mol This problem is somewhat ambiguous because the definition of STP changed in 1982. Prior to 1982, the definition was 273.15 K at a pressure of 1 atmosphere (101325 Pascals). Since 1982, the definition is 273.15 K at a pressure of exactly 100000 Pascals). Because of those 2 different definitions, the volume of 1 mole of gas is either 22.414 Liters (pre 1982 definition), or 22.71098 liters (post 1982 definition). And finally, there's entirely too many text books out there that still use the 35 year obsolete definition. So let's solve this problem using both definitions and you need to pick the correct answer for the text book you're using. First, determine how many moles of gas you have. Just simply divide the volume you have by the molar volume. Pre-1982: 2.1 / 22.414 = 0.093691443 moles Post-1982: 2.1 / 22.71098 = 0.092466287 moles Now determine the molar mass. Simply divide the mass by the moles. So Pre-1982: 3.5 g / 0.093691443 moles = 37.35666667 g/mol Post-1982: 3.5 g / 0.092466287 moles = 37.85163333 g/mol Finally, round to 2 significant figures. So Pre-1982: 37 g/mol Post-1982: 38 g/mol</span>
5 0
3 years ago
Balance this chemical equation with an explanation please: CH3CH2OH + O2 = CO2 + H2O
STALIN [3.7K]
CH3CH2OH + 3O2 = 2CO2 + 3H2O

Basically you do trial and error on both sides so they can be equal
3 0
3 years ago
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