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andreev551 [17]
3 years ago
9

Determine the volume of methane (ch 4 ) gas needed to react completely with 0.660 l of o 2 gas to form methanol (ch 3 oh).

Chemistry
1 answer:
Sloan [31]3 years ago
4 0
Balance Chemical Equation for this reaction is,

                           2 CH₄  +  O₂   →   2CH₃OH

According to this eq, 22.4 L (1 moles) of Oxygen requires 44.8 L (2 mole) CH₄ for complete reaction.
So, the volume of CH₄ required to consume 0.66 L of O₂ is calculated as,

 22.4 L O₂ required to consume  =  44.8 L CH₄
0.660 L O₂ will require                =  X L of CH₄

Solving for X,
                                                X  =  (44.8 L × 0.660 L) ÷ 22.4 L

                                                X  =  1.320 L of CH₄

Result:
            1.320 L of CH
₄ <span>gas is needed to react completely with 0.660 L of O</span>₂<span> gas to form methanol (CH</span>₃OH<span>).</span>
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3.04 Quiz: Chemical Reactions 1 to 10 question
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1. Chemical reactions

2.Substances

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7. Color

8. Gas

Explanation:

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5 0
3 years ago
Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔
Vilka [71]

Answer : The correct option is, (C) 1.1

Solution :  Given,

Initial moles of N_2O_4 = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration N_2O_4.

\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}

\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M

The given equilibrium reaction is,

                           N_2O_4(g)\rightleftharpoons 2NO_2(g)

Initially                      c                 0

At equilibrium   (c-c\alpha)           2c\alpha

The expression of K_c will be,

K_c=\frac{[NO_2]^2}{[N_2O_4]}

K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}

where,

\alpha = degree of dissociation = 40 % = 0.4

Now put all the given values in the above expression, we get:

K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}

K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}

K_c=1.066\aprrox 1.1

Therefore, the value of equilibrium constant for this reaction is, 1.1

4 0
3 years ago
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