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3 years ago

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The number of moles of hydrogen that is needed to produce 0.253 mol of water is...

1 answer:

Oksi-84 [34.3K]3 years ago

8 0

The chemical formula of water is written as H2O we can just use the ratio of the atoms since we are not given the reaction. Hydrogen is a diatomic substance so that it exists as H2. We calculate as follows:

0.253 mol H2O ( 1 mol H2 / 1 mol H2O ) = 0.253 H2 needed

Hope this answers the question. Have a nice day.

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

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Answer:

The answer is " $= 0.078 \ kg \ H_2$ ".

Explanation:

calculating the moles in $CH_4 =\frac{PV}{RT}$

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

Eqution:

$CH_4 +H_2O \to 3H_2+ CO \ (g)$

Calculating the amount of $H_2$ produced:

$= 12.9 \ mol CH_4 \times \frac{3 \ mol \ H_2 }{1 \ mol \ CH_4}\times \frac{2.016 g H_2}{1 \ mol \ H_2}\\\\= 78 \ g \ H_2 \\\\= 0.078 \ kg \ H_2$

So, the amount of dihydrogen produced = $0.078 \frac{kg}{s}$

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What mass of oxygen (O2) forms in a reaction that forms 15.90 g C6H12O6? (Molar mass of O2 = 32.00 g/mol; molar mass of C6H12O6

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The answer is: the mass of oxygen is 16.95 grams.

The overall balanced photosynthesis reaction:

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n(C₆H₁₂O₆) = m(C₆H₁₂O₆) ÷ M(C₆H₁₂O₆).

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n(O₂) = 6 · 0.088 mol.

n(O₂) = 0.53 mol; amount of oxygen.

m(O₂) = 0.53 mol · 32.00 g/mol.

m(O₂) = 16.95 g; mass of oxygen.

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