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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 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.

Answer :

Answer:

The answer is "[tex]= 0.078 \ kg \ H_2[/tex]".

Explanation:

calculating the moles in [tex]CH_4 =\frac{PV}{RT}[/tex]

                                                [tex]=\frac{(0.58 \ atm) \times (923 \ L) }{ (0.0821 \frac{L \cdot atm}{K \cdot mol})(232^{\circ} C +273)}\\\\=\frac{(535.34 \ atm \cdot \ L) }{ (0.0821 \frac{L \cdot atm}{K \cdot mol})(505)K}\\\\=\frac{(535.34 \ atm \cdot \ L) }{ (41.4605 \frac{L \cdot atm}{mol})}\\\\= 12.9 \ mol[/tex]

Eqution:

[tex]CH_4 +H_2O \to 3H_2+ CO \ (g)[/tex]

Calculating the amount of [tex]H_2[/tex] produced:

[tex]= 12.9 \ mol CH_4 \times \frac{3 \ mol \ H_2 }{1 \ mol \ CH_4}\times \frac{2.016 g H_2}{1 \ mol \ H_2}\\\\= 78 \ g \ H_2 \\\\= 0.078 \ kg \ H_2[/tex]

So, the amount of dihydrogen produced = [tex]0.078 \frac{kg}{s}[/tex]

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