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Mashutka [201]
3 years ago
6

How many moles react with 11 moles of oxygen

Chemistry
2 answers:
Lunna [17]3 years ago
7 0
It's one mole. The equation is balanced. You need to look at the coefficient in front of butane ( C4H10 ).
Zinaida [17]3 years ago
6 0
I say it’s 9 moles , hope this is right
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A molecule is broken down into its constituent atoms. Do these atoms have the same properties as the molecule?
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3 years ago
Pulverized coal pellets, which may be approximated as carbon spheres of radius ro= 1 mm, are burned in a pure oxygen atmosphere
Ganezh [65]

Answer:

Explanation:

SO; If we assume that:

P should be the diffusion of oxygen towards the surface ; &

Q should be the  diffusion of carbondioxide away from the surface.

Then  the total molar flux of oxygen is illustrated by :

Na,x = - cD_{PQ}\frac{dy_P}{dr} +y_P(NP,x + N_Q,x)

where;

r is the radial distance from the center of the carbon particle.

Since ;

N_P,x = - N_Q, x ; we have:

Na,x = - cD_{PQ}\frac{dy_P}{dr}

The system is not steady state and the molar flux is not independent of r because the area of mass transfer 4\pi r^{2} is not a constant term.

Therefore,  using quasi steady state assumption, the mass transfer rate   4\pi r^{2}N_{P,x} is assumed to be independent of r at any instant of time.

W_{P}=4\pi r^{2}N_{P,x}

W_{P}=-4\pi r^{2}cD_{PQ}\frac{dy_{P}}{dr}        

       = constant

The oxygen concentration at the surface of the coal particle yP,R will be calculated from the reaction at the surface.

The mole fraction of oxygen at a location far from pellet is 1.

Thus, separating the variables and integrating result into  the following:

W_{P}\int_{R}^{\infty} \frac{dr}{r^{2}}=-4\pi

r^{2}cD_{PQ}\int_{y_{P,R}}^{y_{P,\infty }}dy_{P}

-W_{P}\frac{1}{r}\mid ^{\infty }_{R}= -4\pi cD_{PQ}(y_{P,\infty }-y_{P,R})

=> W_{P}= - 4\pi cD_{PQ}(1-y_{P,R})R

The mole of oxygen arrived at the carbon surface is equal to the mole of oxygen consumed by the chemical reaction.

W_{P} = 4 \pi R^2R"

W_{P}= 4\pi R^{2}k_{1}"C_{O_{2}}\mid _{R}

W_{P}= 4\pi R^{2}k_{1}"c y _{P,R}

-4\pi cD_{PQ}(1-y_{P,R})R= - 4\pi R^{2}k_{1}"c y _{P,R}

y_{P,R}=\frac{D_{PQ}}{D_{PQ}+Rk_{1}}

y_{P,R}=\frac{1.7 \times 10^{-4}}{1.7\times 10^{-4}+10^{-3}\times 0.1}

\mathbf{= 0.631}

Obtaining the total gas concentration from the ideal gas law; we have the following:

where;

R= 0.082m^3atm/kmolK

c=\frac{P}{RT} \\ \\ c=\frac{1}{0.082\times 1450} \\ \\ = 0.008405kmol/m^3

The steady state O_2 molar consumption rate is:

W_{P}= -4\pi cD_{PQ}(1-y_{P,R})R

W_{P}= -4\pi (0.008405)(1.7\times 10^{-4})(1-0.631)(10^{-3})

W_{P}= - 6.66\times 10^{-9}kmol/s

5 0
3 years ago
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