Chemistry, 31.07.2019 18:30 katlynnschmolke

The reform reaction between steam and gaseous methane ( ch4 ) produces "synthesis gas," a mixture of carbon monoxide gas and dihydrogen 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 159. liters per second of methane are consumed when the reaction is run at 294.°c and 0.86atm . calculate the rate at which dihydrogen is being produced. give your answer in kilograms per second. be sure your answer has the correct number of significant digits.

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Answer is: rate of production of hydrogen is 0,01758 kg/s.
Chemical reaction: CH₄ + H₂O → CO + 3H₂.
V(CH₄) = 159 L.
p = 0,86 atm · 101,325 kPa/atm = 87,14 kPa.
T = 294°C = 567 K.
R = 8,314 J/K·mol.
Ideal gas law: p·V = n·R·T.
n(CH₄) = p·V ÷ R·T.
n(CH₄) = 87,14 kPa · 159 L ÷ 8,314 J/K·mol · 567 K.
n(CH₄) = 2,93 mol.
From chemical reaction: n(CH₄) : n(H₂) = 1 : 3.
n(H₂) = 2,93 mol · 3 = 8,79 mol.
m(H₂) = 8,79 mol · 2 g/mol.
m(H₂) = 17,58 g · 0,001 kg/g = 0,01758 kg.

Answer from: Smartpotato9555

The rate of the dihydrogen production is 0.0175 kg/sec for the given reaction.

The given reaction

$\bold {CH_4 + H_2O \rightarrow CO + 3H_2}$

One mole of Methane is react with one mole of water to form 1 mole of CO and 3 moles of Hydrogen.

The rate of hydrogen production can be calculated from ideal gas law

$\bold {pV = nRT}$

Where,

p- pressure = 0,86 atm · 101,325 kPa/atm = 87,14 kPa.

V- volume of methane = 159 L

T - temperature= 294°C = 567 K.

R - gas constant = 8,314 J/K·mol.

n - number of moles = ?

Put the values in the formula, sole it for n,

$\bold{n(CH₄) = \dfrac {pV} {RT}}\\\\\bold {n(CH₄) = \dfrac {87,14 kPa \times 159 L} {8,314 J/K.mol \times 567 K}.}\\\\\bold {n(CH₄) = 2,93 mol}$

Since, the molar ratio of methane to hydrogen is 3:1.

SO,

$\bold {n(H_2) = 2,93 mol \times 3 }\\\\\bold {n (H_2) = 8,79 mol.}\\$

Hence, the mass of the dihydrogen produced,

$\bold{ m(H_2) = 8.79 mol \times 2 g/mol.}\\\\\bold {m(H_2) = 17.58 g \times 0.001 kg/g}\\\\\bold {m(H_2) = 0.01758 kg}$

Therefore, the rate of the dihydrogen production is 0.0175 kg/sec for the given reaction.

To know more about Methane reaction:

link

Answer from: Quest

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.

Answer from: Quest

can you list the answers so i can answer