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Murrr4er [49]
4 years ago
11

The reform reaction between steam and gaseous methane (CH4) produces "synthesis gas," a mixture of carbon monoxide gas and dihyd

rogen gas. Synthesis gas is one of the most widely used industrial chemicals, and is the major industrial source of hydrogen.
Suppose a chemical engineer studying a new catalyst for the reform reaction finds that 924. liters per second of methane are consumed when the reaction is run at 261.°C and 0.96atm. Calculate the rate at which dihydrogen is being produced. Give your answer in kilograms per second. Round your answer to 2 significant digits.
Chemistry
1 answer:
barxatty [35]4 years ago
5 0

Answer:

The rate at which dihydrogen is being produced is 0.12 kg/sec.

Explanation:

CH_4+H_2O\rightarrow CO+3H_2 Haber reaction

Volume of methane consumed in a second = 924 L

Temperature at which reaction is carried out,T= 261°C = 538.15 K

Pressure at which reaction is carried out, P = 0.96 atm

Let the moles of methane be n.

Using an Ideal gas equation:

PV=nRT

n=\frac{PV}{RT}=\frac{0.96 atm\times 924 L}{0.0821 atm l/mol K\times 538.15 K}=20.0769 mol

According to reaction , 1 mol of methane gas produces 3 moles of dihydrogen gas.

Then 20.0769 mol of dihydrogen will produce :

\frac{3}{1}\times 20.0769 mol=60.2307 mol of dihydrogen

Mass of 24.3194 moles of ammonia =24.3194 mol × 2 g/mol

=120.46 g=0.12046 kg ≈ 0.12 kg

924 L of methane are consumed in 1 second to produce 0.12 kg of dihydrogen in 1 second. So the rate at which dihydrogen is being produced is 0.12 kg/sec.

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calculate the water potential of a solution of 0.15m sucrose. the solution is at standard temperature.
Mrac [35]

Answer:

The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.

Explanation:

Water potential = Pressure potential + solute potential

P_w=P_p+P_s

P_w=P_p+(-iCRT)

We have :

C = 0.15 M, T = 273.15 K

i = 1

The water potential of a solution of 0.15 m sucrose= P_w

P_p=0 bar (At standard temperature)

P_s=-iCRT=-\times 1\times 8.314\times 10^{-2}bar L/mol K\times 273.15 K=-3.406 bar

P_w=0 bar+(-3.406 ) bar

The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.

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3 years ago
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Answer:

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Explanation:

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Answer:

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Explanation:

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