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Murrr4er [49]
3 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]3 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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The following reactions can be used to prepare samples of metals. Determine the enthalpy change under standard state conditions
mamaluj [8]

Answer:

a) 62.1 kJ/mol

b) 2.82 kJ/mol

c) 270.91 kJ/mol

d) -851.5 kJ/mol

Explanation:

The enthalpy change for a reaction in standard conditions (ΔH°rxn) can be calculated by:

ΔH°rxn = ∑n*ΔH°f, products - ∑n*ΔH°f, reagents

Where n is the number of moles in the stoichiometry reaction, and ΔH°f is the enthalpy of formation at standard conditions. ΔH°f = 0 for substances formed by only a single element. The values can be found in thermodynamics tables.

a) 2Ag₂O(s) → 4Ag(s) + O₂(g)

ΔH°f, Ag₂O(s) = -31.05 kJ/mol

ΔH°rxn = 0 - (2*(-31.05)) = 62.1 kJ/mol

b) SnO(s) + CO(g) → Sn(s) + CO₂(g)

ΔH°f,SnO(s) = -285.8 kJ/mol

ΔH°f,CO(g) = -110.53 kJ/mol

ΔH°f,CO₂(g) = -393.51 kJ/mol

ΔH°rxn = [-393.51] - [-110.53 - 285.8] = 2.82 kJ/mol

c) Cr₂O₃(s) + 3H₂(g) → 2Cr(s) + 3H₂O(l)

ΔH°f,Cr₂O₃(s) = -1128.4 kJ/mol

ΔH°f,H₂O(l) = -285.83 kJ/mol

ΔH°rxn = [3*(-285.83)] - [( -1128.4)] = 270.91 kJ/mol

d) 2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s)

ΔH°f,Fe₂O₃(s) = -824.2 kJ/mol

ΔH°f,Al₂O₃(s) = -1675.7 kJ/mol

ΔH°rxn = [-1675.7] - [-824.2] = -851.5 kJ/mol

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valina [46]

Answer:

They have properties of both metals and nonmetals

Explanation:

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Answer:

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