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fenix001 [56]
3 years ago
15

Ammonia is produced by the following reaction. 3H2(g) N2(g) Right arrow. 2NH3(g) When 7. 00 g of hydrogen react with 70. 0 g of

nitrogen, hydrogen is considered the limiting reactant because 7. 5 mol of hydrogen would be needed to consume the available nitrogen. 7. 5 mol of nitrogen would be needed to consume the available hydrogen. Hydrogen would produce 7. 5 mol more ammonia than nitrogen. Nitrogen would produce 7. 5 mol more ammonia than hydrogen.
Chemistry
1 answer:
harkovskaia [24]3 years ago
6 0

In the ammonia production process given by the reaction 3H₂(g) + N₂(g) → 2NH₃(g), when 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because <u>7.5 moles of hydrogen would be needed to consume the available nitrogen</u> (option 1).

The reaction is the following:

3H₂(g) + N₂(g) → 2NH₃(g)   (1)

To know why hydrogen is considered the limiting reactant, we need to calculate the number of moles of nitrogen and hydrogen with the following equation:

n = \frac{m}{M}

Where:    

m: is the mass

M: is the molar mass

  • For <em>hydrogen </em>we have:

n_{H_{2}} = \frac{m}{M} = \frac{7.00 g}{2.016 g/mol} = 3.47 \:moles

  • And for <em>nitrogen</em>:

n_{N_{2}} = \frac{m}{M} = \frac{70.0 g}{28.013 g/mol} = 2.50 \:moles

We can see in reaction (1) that <u>3 moles of hydrogen</u> react with <u>1 mol of nitrogen</u>, so the number of hydrogen moles needed to react nitrogen is:

n_{H_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*n_{N_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*2.50 \:moles = 7.50 \:moles

Since we have <u>3.47 moles of hydrogen</u> and we need <u>7.50 moles</u> to react with all the mass of nitrogen, the <em>limiting reactant</em> is <em>hydrogen</em>.

We can find the number of ammonia moles produced with the limiting reactant (hydrogen) konwing that <u>3 moles of hydrogen</u> produces <u>2 moles of ammonia</u>, so:

n_{NH_{3}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*n_{H_{2}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*3.47 \:moles = 2.31 \:moles

Hence, hydrogen would produce <u>2.31 moles of ammonia</u>.

Therefore, hydrogen is the limiting reactant because <u>7.5 moles of hydrogen would be needed to consume the available nitrogen</u> (option 1).

Find more about limiting reactants here:

brainly.com/question/2948214?referrer=searchResults

   

I hope it helps you!                        

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The number of mole will be 65.81 mole.

An ideal gas would be one for which both the overall volume of the molecules and even the forces that exist between them are so negligible as to have no influence on the behavior of something like the gas.

Number of ideal gas can be calculated by using the formula:

PV = nRT

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V= 1750 dm^{3} = 1750 L

P = 125,000 p = 1.2 atm

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T = 273+127 = 400 K

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1.23atm x 1750L = n x 0.0820atm x Liter/ mole x kelvin  x 400K

n = 65.81 mole.

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2 years ago
Concentration of 10.00 mL of HBr if it takes 5 mL of a 0.253 M LiOH solution to<br> neutralize it?
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In a titration, for an acid to neutralize a base, at the equivalence point, there should be an equal number of moles of H+ and OH-.

Moles of OH- can be found by multiplying the concentration of the base by the volume. (You will need to keep in mind the stoichimetric coefficients if the strong base is Ca(OH)₂, Ba(OH)₂, or Sr(OH)₂.

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In the Haber reaction, patented by German chemist Fritz Haber in 1908, dinitrogen gas combines with dihydrogen gas to produce ga
VashaNatasha [74]

The question is incomplete, complete question is :

In the Haber reaction, patented by German chemist Fritz Haber in 1908, dinitrogen gas combines with dihydrogen gas to produce gaseous ammonia. This reaction is now the first step taken to make most of the world's fertilizer. Suppose a chemical engineer studying a new catalyst for the Haber reaction finds that 348 liters per second of dinitrogen are consumed when the reaction is run at 205°C and 0.72 atm. Calculate the rate at which ammonia is being produced.

Answer:

The rate of production of ammonia is 217.08 grams per second.

Explanation:

N_2+3H_2\rightarrow 2NH_3

Volume of dinitrogen used in a second = 348 L

Temperature of the gas = T = 205°C = 205+273 K = 478 K

Pressure of the gas = P = 0.72 atm

Moles of dinitrogen = n

n=\frac{PV}{RT}=\frac{0.72 atm\times 348 L}{0.0821 atm L/mol K\times 478 K}=6.385 mol

According to reaction, 1 mole of dinitriogen gives 2 mole of ammonia.Then 6.385 moles of dinitrogen will give:

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Mass of 12.769 moles of ammonia;

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217.08 grams of ammonia is produced per second.So, the rate of production of ammonia is 217.08 grams per second.

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3 years ago
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