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padilas [110]
3 years ago
15

Given the following chemical reaction: 2 O2 + CH4 CO2 + 2 H2O What mass of CH4 is required to completely react with 100 grams of

O2? 6 C 12.01 Carbon 1 H 1.01 Hydrogen 8 O 16.00 Oxygen A. 10 grams CH4 B. 20 grams CH4 C. 25 grams CH4 D. 45 grams CH4
Chemistry
1 answer:
AfilCa [17]3 years ago
6 0

Answer:

The mass methane (CH4) required is, 25 grams  (option C)

Explanation:

Mass of oxygen gas = 100 g

Molar mass of oxygen gas = 32 g/mole

Molar mass of methane gas = 16 g/mole

<u>Step 1:</u> Balance the reaction

2O2 + CH4 → CO2 + 2H2O

<u>Step 2:</u> Calculate the moles of O2:

Moles O2 = mass of O2 / Molar mass of o2  

Moles O2 = 100g / (32g/mole) = 3,125 moles

⇒From the balanced reaction we conclude that  2 moles of O2 react with 1 mole of CH4

⇒ So, 3.125 moles of O2 react with 3.125/2 = 1.563 moles of CH4

<u>Step 3:</u> Calculate the mass of CH4:

Mass of CH4 = moles of CH4 x Molar mass of CH4

Mass of CH4 = 1.563 moles / (16g/ moles) = 25.008 grams CH4

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Before any reaction occurs, the concentration of A in the reaction below is 0.0510 M. What is the equilibrium constant if the co
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Answer:

0.0119

Explanation:

There was a part missing. I think this is the whole question:

<em>Before any reaction occurs, the concentration of A in the reaction below is 0.0510 M. What is the equilibrium constant if the concentration of A at equilibrium is 0.0153 M?</em>

A (aq) ⇌ 2B (aq) + C(aq)

<em>Remember to use correct significant figures in your answer. Do not include units in your response.</em>

First, we have to make an ICE Chart, which stands for initial, change and equilibrium. We will call "x" unknown concentrations.

        A (aq) ⇌ 2B (aq) + C (aq)

I       0.0510       0            0

C         -x          +2x          +x

E    0.0510-x     2x            x

Since the concentration at equilibrium of A is 0.0153 M, we get

0.0510 - x = 0.0153 \\x = 0.0357

We can use the value of x to calculate the concentrations at equilibrium.

[A]e = 0.0153 M \\[B]e = 2x = 2(0.0357) = 0.0714 M \\[C]e = x = 0.0357 M \\

The equilibrium constant, Kc, is the ratio of the equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients.

Kc = \frac{[B]^{2}  \times [C]}{[A]} = \frac{0.0714^{2}  \times 0.0357}{0.0153} = 0.0119

The equilibrium constant for this reaction at equilibrium is 0.0119.

You can learn more about equilibrium here: brainly.com/question/4289021

7 0
2 years ago
Read 2 more answers
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