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A mixture of methane and air is capable of being ignited only if the mole percent of methane is between 5% and 15%. A mixture co

DaniilM [7]

Answer:

Explanation:

The air 9% mole% methane have an average molecular weight of:

9%×16,04g/mol + 91%×29g/mol = 27,8g/mol

And a flow of 700000g/h÷27,8g/mol = 25180 mol/h

In the reactor where methane solution and air are mixed:

In = Out

Air balance:

91% air×25180 mol/h + 100% air×X = 95%air×(X+25180)

Where X is the flow rate of air in mol/h = 20144 mol air/h

The air in the product gas is

95%×(20144 + 25180) mol/h = 43058 mol air× 21%O₂ = 9042 mol O₂ ×32g/mol = 289 kg O₂

43058 mol air×29g/mol 1249 kg air

Percent of oxygen is: $\frac{289kg}{1249 kg}$ =0,231 kg O₂/ kg air

I hope it helps!

4 0

3 years ago

a 25.0-ml volume of a sodium hydroxide solution requires 19.6 ml of a 0.189 m hydrochloric acid for neutralization. a 10.0- ml v

Rashid [163]

Concentration of NaOH = 0.148 molar, M

Concentration of H3PO4 = 0.172 molar, M

Concentration x Volume will give the number of moles of solute in that volume. C*V = moles

Concentration has a unit of (moles/liter). When multiplied by the liters of solution used, the result is the number of moles.

Original HCl solution: (0.189 moles/L)*(0.0196 L)= 0.00370 moles of HCl

The neutralization of 25.0 ml of sodium hydroxide, NaOH, requires 0.00370 moles of HCl. The reaction is:

NaOH + HCl > NaCl and H2O

This balanced equation tells us that neutralization of NaOH with HCl requires the same number of moles of each. We just determined that the moles of HCl used was 0.00370 moles. Therefore, the 25.0 ml solution of NaOH had the same number of moles: 0.00370 moles NaOH.

The 0.00370 moles of NaOH was contained in 25.0 ml (0.025 liters). The concentration of NaOH is therefore:

(0.00370 moles of NaOH)/(0.025 L) = 0.148 moles/liter or Molar, M

====

The phosphoric acid problem is handled the same way, but with an added twist. Phosphoric acid is H3PO4. We learn the 34.9 ml of the same NaOH solution (0.148M) is needed to neutralize the H3PO4. But now the acid has three hydrogens that will react. The balanced equation for this reaction is:

H3PO4 + 3NaOH = Na3PO4 + 3H2O

Now we need three times the moles of NaOH to neutralize 1 mole of H3PO4.

The moles of NaOH that were used is:

(0.148M)*(0.0349 liters) = 0.00517 moles of NaOH

Since the molar ratio of NaOH to H3PO4 is 3 for neutralization, the NaOH only neutralized (0.00517)*(1/3)moles of H3PO4 = 0.00172 moles of H3PO4.

The 0.00172 moles of H3PO4 was contained in 10.0 ml. The concentration is therefore:

(0.00172 moles H3PO4)/(0.010 liters H3PO4)

Concentration of H3PO4 = 0.172 molar, M

5 0

1 year ago

Determine the number of 3s electrons in Na .

Scorpion4ik [409]

Answer:

Explanation:The atomic number of sodium is 11. That is, the number of electrons in sodium is 11. Therefore, a sodium atom will have two electrons in the first shell, eight in the 2nd orbit, and an electron in the 3rd shell.

3 0

1 year ago

Why is melting ice cream a physical?

worty [1.4K]

Melting ice cream is a physical change because it is a phase change, the water in the ice cream gains thermal energy and the inter-molecular forces are overcome and the molecules move further apart forming a liquid. It is not a chemical change because no chemical reactions occurred.

3 0

4 years ago

Read 2 more answers

The reactant concentration in a second-order reaction was 0.560 m after 115 s and 3.30×10−2 m after 735 s . what is the rate con

Aloiza [94]

Use the formula for second order reaction:

$\frac{1}{C} = \frac{1}{C_0} + kt$

C = concentration at time t
C0 = initial conc.
k = rate constant
t = time

1st equation : $\frac{1}{0.56} = \frac{1}{C_0} + 115k$

2nd Equation: $\frac{1}{0.033} = \frac{1}{C_0} + 735k$

Find $\frac{1}{C_0}$ from 1st equation and put it in 2nd equation:

$\frac{1}{0.033} = \frac{1}{0.56} - 115k + 735k$

$\frac{1}{0.033} - \frac{1}{0.56} = 620k$

k = 0.046

3 0

3 years ago