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3 years ago

5

Using the balanced equation for the combustion of ethane: 2C2H6 + 7O2 → 4CO2 + 6H2O, how many moles of O2 needed to produce 12 m

oles of H2O?

1 answer:

pickupchik [31]3 years ago

3 0

Answer:

14 moles of oxygen needed to produce 12 moles of H2O.

Explanation:

We are given that balance eqaution

$2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O$

We have to find number of moles of O2 needed to produce 12 moles of H2O.

From given equation

We can see that

6 moles of H2O produced by Oxygen =7 moles

1 mole of H2O produced by Oxygen= $\frac{7}{6}$ moles

12 moles of H2O produced by Oxygen= $\frac{7}{6}\times 12$ moles

12 moles of H2O produced by Oxygen= $7\times 2$ moles

12 moles of H2O produced by Oxygen=14 moles

Hence, 14 moles of oxygen needed to produce 12 moles of H2O.

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A tank contains 115 miles of neon gas. It has a pressure of 57 atm at a temperature of 45 c. Calculate the volume of the tank.

Elena-2011 [213]

Answer:

Volume occupied by Neon gas is 52.67 L

Explanation:

Using Ideal Gas Equation:

PV = nRT

where

P = pressure exerted by the gas = 57 atm

V = volume occupied = ?

n = number of moles = 115 moles

R = Ideal gas constant = 0.0821 L.atm/K.mol

(R value should be taken according to the units of Temperature,pressure, volume and mole)

T = Temperature = 45 + 273 = 318 K

(For temperature conversion from C to K add 273 to temperature:T + 273)

PV = nRT , So

$V = \frac{nRT}{P}$

Put values of T,P,n,R

$V = \frac{115\times 0.0821\times 318}{57}$

$V = \frac{3002.39}{57}$

V = 52.67 L

Volume occupied by 115 moles of Neon gas at 57 atm Pressure and 45 C temperature is 52.67 L

6 0

4 years ago

If the forward reaction is favored in the equilibrium, what will increase?

Lorico [155]

Answer:

Option C:- concentration of products

Explanation:

According to Le Chateleir's principle,

if the forward reaction is favored in the equilibrium then concentration of products will increase and concentration of reactants will decrease.

If the reverse reaction is favored in the equilibrium then concentration of products will decrease and concentration of reactants will increase.

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3 years ago

Given that the approximate bond length of o–h in water is roughly 1 å, what is the approximate distance each proton "hops" betwe

Ksenya-84 [330]

The hydrogen bond is a comparatively weak interaction between a proton hop and an electronegative atom present in a molecule. The hydrogen bond plays a very important role for the determination of state of a compound like gaseous, liquid or solid. The strength of the hydrogen bond depends on the close distance between the participants i.e. the electronegative atom and proton hop. There remains strong hydrogen bond between the two water molecules which is expressed as O....H. The distance of the hydrogen bond is 1.8A° formed between each proton hops of two neighboring water molecules. The hydrogen bond interaction is shown in the figure.

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3 years ago

When writing a PSA, it is not necessary to brainstorm because you already have all of the information from your persuasive essay

Sergio [31]

YES it is very important to brainstorm

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3 years ago

A mixture of A and B is capable of being ignited only if the mole percent of A is 6 %. A mixture containing 9.0 mole% A in B flo

atroni [7]

Explanation:

As it is given that mixture (contains 9 mol % A and 91% B) and it is flowing at a rate of 800 kg/h.

Hence, calculate the molecular weight of the mixture as follows.

Weight = $0.09 \times 16.04 + 0.91 \times 29$

= 27.8336 g/mol

And, molar flow rate of air and mixture is calculated as follows.

$\frac{800}{27.8336}$

= 28.74 kmol/hr

Now, applying component balance as follows.

$0.09 \times 28.74 + 0 \times F_{B} = 0.06F_{p}$

$F_{p}$ = 43.11 kmol/hr

$F_{A} + F_{B} = F_{p}$

$F_{B}$ = 43.11 - 28.74

= 14.37 kmol/hr

So, mass flow rate of pure (B), is $F_{B}$ = $14.37 \times 29$

= 416.73 kg/hr

According to the product stream, 6 mol% A and 94 mol% B is there.

Molecular weight of product stream = $Mol. weight \times 43.11 kmol/hr$

= $0.06 \times 16.04 + 0.94 \times 29$

= 28.22 g/mol

Mass of product stream = 1216.67 kg/hr

Hence, mole of $O_{2}$ into the product stream is as follows.

$0.21 \times 0.94 \times 43.11$

= $8.5099 kmol/hr \times 329 g/mol$

= 272.31 kg/hr

Therefore, calculate the mass % of $O_{2}$ into the stream as follows.

$\frac{272.31}{1216.67} \times 100$

= 22.38%

Thus, we can conclude that the required flow rate of B is 272.31 kg/hr and the percent by mass of $O_{2}$ in the product gas is 22.38%.

4 0

3 years ago