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kvv77 [185]
3 years ago
14

Is the final answer oxidation or reduction?

Chemistry
1 answer:
zhannawk [14.2K]3 years ago
4 0

Answer:

oxidation

Explanation:

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A(g) + 2B(g) → C(g) + D(g)
WITCHER [35]

Answer:

0.169

Explanation:

Let's consider the following reaction.

A(g) + 2B(g) ⇄ C(g) + D(g)

We can find the pressures at equilibrium using an ICE chart.

       A(g) + 2 B(g) ⇄ C(g) + D(g)

I       1.00     1.00        0        0

C       -x        -2x        +x       +x

E    1.00-x  1.00-2x     x         x

The pressure at equilibrium of C is 0.211 atm, so x = 0.211.

The pressures at equilibrium are:

pA = 1.00-x = 1.00-0.211 = 0.789 atm

pB = 1.00-2x = 1.00-2(0.211) = 0.578 atm

pC = x = 0.211 atm

pD = x = 0.211 atm

The pressure equilibrium constant (Kp) is:

Kp = pC × pD / pA × pB²

Kp = 0.211 × 0.211 / 0.789 × 0.578²

Kp = 0.169

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Which of the following is a characteristic of alcohols
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Gaseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 9.6 g of methane is
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Answer:

21.6 g

Explanation:

The reaction that takes place is:

  • CH₄ + 2O₂ → CO₂ + 2H₂O

First we<u> convert the given masses of both reactants into moles</u>, using their <em>respective molar masses</em>:

  • 9.6 g CH₄ ÷ 16 g/mol = 0.6 mol CH₄
  • 64.9 g O₂ ÷ 32 g/mol = 2.03 mol O₂

0.6 moles of CH₄ would react completely with (2 * 0.6) 1.2 moles of O₂. As there are more O₂ moles than required, O₂ is the reactant in excess and CH₄ is the limiting reactant.

Now we <u>calculate how many moles of water are produced</u>, using the <em>number of moles of the limiting reactant</em>:

  • 0.6 mol CH₄ * \frac{2molH_2O}{1molCH_4} = 1.2 mol H₂O

Finally we<u> convert 1.2 moles of water into grams</u>, using its <em>molar mass</em>:

  • 1.2 mol * 18 g/mol = 21.6 g
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