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3 years ago

N2-3H2 → 2NH3

Chemistry

1 answer:

aniked [119]3 years ago

8 0

Answer:

6 mols H2 are needed

Explanation:

N2 = 28.01g/mol

H2 = 2.02g/mol

$\frac{2 mol N_{2} }{1}$ * $\frac{3 mol H_{2} }{ 1 mol N_{2} }$ = 6 mol H2

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A gas container has a volume of 446.9 with a temp of 14c. When the volume is decreased to 238.7l the new temp is what

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Answer:

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Be sure to answer all parts. Hydrogen iodide decomposes according to the reaction 2 HI(g) ⇌ H2(g) + I2(g) A sealed 1.50−L contai

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Answer : The concentration of $HI$ and $I_2$ at equilibrium is, 0.0158 M and 0.00302 M respectively.

Explanation :

First we have to calculate the concentration of $H_2, I_2\text{ and }HI$

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Now we have to calculate the value of equilibrium constant (K).

The given chemical reaction is:

$2HI(g)\rightleftharpoons H_2(g)+I_2(g)$

Initial conc. 0.0163 0.00415 0.00276

At eqm. (0.0163-2x) (0.00415+x) (0.00276+x)

As we are given:

Concentration of $H_2$ at equilibrium = 0.00467 M

That means,

(0.00415+x) = 0.00467

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Concentration of $I_2$ at equilibrium = (0.00276+x) = (0.00276+0.00026) = 0.00302 M

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3 years ago

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