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2 years ago

N2 + 3H2, -> 2NH3 If I have 50.6 L of N2 and excess H2, how many liters of NH3 can I produce?

Chemistry

1 answer:

Anastaziya [24]2 years ago

3 0

Answer:

C. 101.2 L

Explanation:

N2 + H2= NH3

Balancing it,

N2 + 3 H2 = 2.NH3

(1 mol) (3 mol) (2 mol)

which means

1 molecule of nitrogen reacts with 3 molecule of hydrogen to form ammonia.

Likewise,

50.6 l of nitrogen reacts with 50.6 × 3= 151.8 l of hydrogrn to form 50.6 × 2= 101.2 l of ammonia.

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Answer:

D.) The nuclear model describes the atom as a dense, positive nucleus surrounded by a cloud of negative electrons.

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2 years ago

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Consider the reaction for the decomposition of hydrogen disulfide: 2H2S(g)⇌2H2(g)+S2(g), Kc = 1.67×10−7 at 800∘C A 0.500 L react

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Answer:

Molar concentration of S₂ is 1.77×10⁻⁶M

Explanation:

For the reaction:

2H₂S(g) ⇄ 2H₂(g) + S₂(g)

The equilibirum constant, K, is defined as:

$K = \frac{[S_2][H_2]^2}{[H_2S]^2}$ (1)

Concentrations in equilibirum are:

[H₂S] : 0,163/0.500L - X

[H₂] : 0,0500/0.500L + X

[S₂] : X

Replacing the concentrations and the equilibrium value in (1):

$K = \frac{[X][0.1+X]^2}{[0326-X]^2}$

1.67x10⁻⁷ = X (X² + 0.2X + 0.01) / (X² -0.652X + 0.106)

1.67x10⁻⁷X² - 1.09x10⁻⁷X + 1.77x10⁻⁸ = X³ + 0.2X² + 0.01X

0 = X³ + 0.2X² + 0.01X - 1.77x10⁻⁸

Solving for X:

X = 1.77×10⁻⁶

As [S₂] = X, molar concentration of S₂ is 1.77×10⁻⁶M

I hope it helps!

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3 years ago

HI decomposes to H2 and I2 by the following equation: 2HI(g) → H2(g) + I2(g);Kc = 1.6 × 10−3 at 25∘C If 1.0 M HI is placed into

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Answer: The concentration of hydrogen gas at equilibrium is 0.037 M

Explanation:

We are given:

Initial concentration of HI = 1.0 M

The given chemical equation follows:

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

Initial: 1.0

At eqllm: 1.0-2x x x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[H_2][I_2]}{[HI]^2}$

We are given:

$Kc=1.6\times 10^{-3}$

Putting values in above expression, we get:

$1.6\times 10^{-3}=\frac{x\times x}{(1.0-2x)^2}\\\\x=-0.043,0.037$

Neglecting the negative value of 'x' because concentration cannot be negative

So, equilibrium concentration of hydrogen gas = x = 0.037 M

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