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4 years ago

How many atoms are in 15.0 moles of C2H6O

Chemistry

2 answers:

MA_775_DIABLO [31]4 years ago

6 0

Answer: $813.1\times 10^{23}$ atoms.

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to the molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

1 molecule of $C_2H_6O$ contains = 9 atoms

1 mole of $C_2H_6O$ contains = $9 \times 6.023\times 10^{23}=54.21\times 10^{23}$ atoms

Thus 15 moles $C_2H_6O$ contains = $\frac{54.21\times 10^{23}}{1}\times 15=813.1\times 10^{23}$ atoms.

Thus the answer is $813.1\times 10^{23}$ atoms.

Burka [1]4 years ago

3 0

1 mole = 6.22 x 10^23 molecules (Avogadro's number)
15 moles x (6.22 x 10^23) = 9.33 x 10^24 atoms

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The question is incomplete. The complete question is :

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$CH_3(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g ) \ \ \ \ \ \ \ \ \ \ K_1$

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$CH_3(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g ) \ \ \ \ \ \ \ \ \ \ K_1$

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$CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g) \ \ \ \ \ \ \ \ \ \ \ \ K_2$

$K_2 = \frac{[CO_2][H_2]}{[CO][H_2O]}$ ...................(2)

$CH_4(g) + 2H_2O(g) \rightleftharpoons CO_2(g) + 4H_2(g) \ \ \ \ \ \ \ \ \ K$

$K=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}$

On multiplication of equation (1) and (2), we get

$K_1 \times K_2=\frac{[CO][H_2]^3}{[CH_4][H_2O]} \times \frac{[CO_2][H_2]}{[CO][H_2O]}$

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Comparing equation (3) and equation (4), we get

$K=K_1K_2$

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