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3 years ago

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Coal containing 15.0% H2O, 2.0% S and 83.0% C by mass is burnt with the stoichiometric amount of air in a furnace. What is the m

ole fraction of SO2(g) in the exhaust

1 answer:

devlian [24]3 years ago

5 0

Answer:

This is a coal combustion process and we will assume

Inlet coal amount = 100kg

It means that there are

15kg of H2O, 2kg of Sulphur and 83kg of Carbon

Now to find the mole fraction of SO2(g) in the exhaust?

Molar mass of S = 32kg/kmol

Initial moles n of S = 2/32 = 0.0625kmols

Reaction: S + O₂ = SO₂

That is 1 mole of S reacts with 1 mole of O₂ to give 1 mole of SO₂

Then, it means for 0.0625 kmoles of S, we will have 0.0625 kmole of SO2 coming out of the exhaust

The mole fraction of SO2(g) in the exhaust=0.0625kmols

Explanation:

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Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 42. g of butane is m

taurus [48]

Answer:

127 grams of carbon dioxide

Explanation:

We need to determine the chemical equation first. Butane has a chemical formula of $C_4H_{10}$ , oxygen is $O_2$ , carbon dioxide is $CO_2$ , and water is $H_2O$ . The reactants are butane and oxygen and the products are carbon dioxide and water. So we write:

$C_4H_{10}+O_2$ ⇒ $CO_2+H_2O$

But remember! We need to balance this. Currently, there are 4 carbon atoms (C), 10 hydrogen atoms (H), and 2 oxygen atoms (O) on the left, while there are 1 carbon atom (C), 2 hydrogen atoms (H), and 3 oxygen atoms (O) on the right. Let's place a coefficient of 4 in front of the carbon dioxide and a coefficient of 5 on the water, so that we have equal numbers of carbon and hydrogen atoms on each side:

$C_4H_{10}+O_2$ ⇒ $4CO_2+5H_2O$

However, we need to ensure that there are equal numbers of O atoms, as well. On the left, we have 2 and on the right we have 13, so let's put a coefficient of 6.5 on the oxygen:

$C_4H_{10}+6.5O_2$ ⇒ $4CO_2+5H_2O$

Finally, multiply everything by 2 to get whole number coefficients:

$2C_4H_{10}+13O_2$ ⇒ $8CO_2+10H_2O$

Ah, now we can actually get to the problem!

We need to determine the limiting reactant, so let's convert the 42 g of butane and 150 g of oxygen into moles of any product, say, carbon dioxide. To convert to moles, we need to find the molar mass of each compound.

The molar mass of butane is 4 * 12.01 + 10 * 1.01 = 58.14 g/mol, while the molar mass of oxygen is 2 * 16 = 32 g/mol. We can now set up the equations:

$42 gC_4H_{10}*\frac{1molC_4H_{10}}{58.14gC_4H_{10}} *\frac{8molCO_2}{2molC_4H_{10}} =2.8896molCO_2$

$150 gO_2*\frac{1molO_2}{32gO_2} *\frac{8molCO_2}{13molO_2} =2.8846molCO_2$

Clearly, we see that 2.8846 < 2.8896, which means that oxygen is the limiting reactant. In other words, the most products can be made when the oxygen is all used up.

Now let's finally convert moles of carbon dioxide into grams by multiplying by its molar mass, which is 12.01 + 2 * 16 = 44.01 g/mol:

$2.8846molCO_2*\frac{44.01gCO_2}{1molCO_2} =127gCO_2$

Notice that we have 3 significant figures because we had 3 significant figures at the start with 150. grams of oxygen.

<em>~ an aesthetics lover</em>

4 0

3 years ago

Your friend had an upset stomach caused by indigestion. His mother explained to him

ddd [48]

Answer:

because it may cause burn that causes acid to build up which causes you to keep having an upset stomach

4 0

2 years ago

Critical Thinking: How are Volcanoes connected to the formation of the atmosphere and oceans????

garri49 [273]

Not sure how they're connected to atmosphere other than polluting it with hot ash, but im fairly certain it creates land masses which separate oceans and form land underwater with certain underwater volcanoes

5 0

3 years ago

Examine the images.

gavmur [86]

Answer:

4 th image

Explanation:

water is being boiled n the 4th image indicating that the change between liquids and gasses is water vapor.

5 0

3 years ago

Read 2 more answers

Consider the reaction of peroxydisulfate ion (S2O2−8) with iodide ion (I−) in aqueous solution: S2O2−8(aq)+3I−(aq)→2SO2−4(aq)+I−

kolbaska11 [484]

Answer:

r = 3.61x $10^{-6}$ M/s

Explanation:

The rate of disappearance (r) is given by the multiplication of the concentrations of the reagents, each one raised of the coefficient of the reaction.

r = k. $[S2O2^{-8} ]^{x} x [I^{-} ]^{y}$

K is the constant of the reaction, and doesn't depends on the concentrations. First, let's find the coefficients x and y. Let's use the first and the second experiments, and lets divide 1º by 2º :

$\frac{r1}{r2} = \frac{0.018^{x} x0.036^{y} }{0.027^xx0.036^y}$

$\frac{2.6x10^{-6}}{3.9x10^{-6}} = (\frac{0.018}{0.027})^xx(\frac{0.036}{0.036})^y$

$0.67 = 0.67^x$

x = 1

Now, to find the coefficient y let's do the same for the experiments 1 and 3:

$\frac{r1}{r3} = \frac{0.018x0.036^y}{0.036x0.054^y}$

$\frac{2.6x10^{-6}}{7.8x10^{-6}} = (\frac{0.018}{0.036})x(\frac{0.036}{0.054})^y$

$0.33 = 0.5x 0.67^y$

$0.67 = 0.67^y$

y = 1

Now, we need to calculate the constant k in whatever experiment. Using the first :

$2.6x10^{-6} = kx0.018x0.036$ $kx6.48x10^{-4} = 2.6x10^{-6}$

k = 4.01x10^{-3} M^{-1}s^{-1}[/tex]

Using the data given,

r = $4.01x10^{-3}x1.8x10^{-2}x5.0x10^{-2}$

r = 3.61x $10^{-6}$ M/s

7 0

3 years ago