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2 years ago

Plllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllls help me

Chemistry

1 answer:

larisa [96]2 years ago

6 0

Answer:

Explanation:

equilibrium constant

Kc = [ C ]² / [ A ] [ B ]

= .5² / .2 x 3

= .4167

Let moles of A to be added be n

concentration of A unreacted becomes .2 + n M

increase of product C by .2 M will require use of A and B be .1 M

So unreacted A = .2 + n - .1 = n + .1

Kc = [ C ]² / [ A ] [ B ]

.4167 = .7² / ( n + .1 ) ( 3 - .1 )

n + .1 = .4

n = . 3 moles .

So .3 moles of A to be added .

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First thing is to find the numbers of moles of each atom.

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$=71.1 \ g\times \dfrac{1 \ mol }{1.008 \ g} \\ \\ =4.44 \ mol \ of \ oxygen$

Now; we use the smallest no of moles to divide the respective moles from above.

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$\dfrac{2.22 \ mol \ of \ carbon}{2.22} =1 \ mol \ of \ hydrogen$

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