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Artyom0805 [142]
3 years ago
11

1 mol of hydrogen gas and 1 mol of iodine vapour were mixed and allowed to react. After t seconds 0.8 mol of hydrogen remained.

Chemistry
1 answer:
vlada-n [284]3 years ago
8 0
At the beginning there was just 1 mole of hydrogen and 1 mole of iodin.
The reaction goes:
H_2+I_2\Rightarrow\ 2HI
One mole of hydrogen goes with one mole of iodin, creating 2 moles of product.
If 0.8 of hydrogen remaind, it means that 1-0.8=0.2 reacted.
So:
1 mole H2 react with 1 mole I2 creating 2 moles of products.
So 0.2 of hydrogen reacted with 0.2 mole of iodin creating 0.4 moles.
1\ mole\ H_2\ ---\ 2\ moles\ HI\\
0.2\ mole\ \ \ ---\ x\\
x=0.4
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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:

\frac{2}{1}\times 6.385 mol=12.769 mol

Mass of 12.769 moles of ammonia;

12.769 mol 17 g/mol = 217.08 g

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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4 years ago
Given the balanced equation below, calculate the moles of aluminum that are needed to react completely with 13.2 moles of FeO?
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The number of moles of aluminium  that are needed to react completely  with 13.2 moles of FeO  is  8.8 moles



calculation


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= 13.2  x  2/3=8.8  moles  of Al

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III. The catalyst speeds up the reaction.

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