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3 years ago

Given the equation 2CH4 S8 --> 2CS2 4H2S Calculate the moles of H2S produced when 1.5 mol S8 is used.

Chemistry

1 answer:

mafiozo [28]3 years ago

8 0

Answer:

6.0 moles of H₂S are produced.

Explanation:

Let's consider the following balanced equation.

2 CH₄ + S₈ → 2 CS₂ + 4 H₂S

The molar ratio of S₈ to H₂S is 1 mol S₈: 4 mol H₂S. The moles of H₂S produced when 1.5 mol of S₈ react are:

1.5 mol S₈ × (4 mol H₂S/ 1 mol S₈) = 6.0 mol H₂S

6.0 moles of H₂S are produced, when 1.5 mol of S₈ react.

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2 years ago

Kc = 3.07 x 10-4 at 24°C for 2NOBr(g) ↔ 2NO(g) + Br2(g). If the initial concentration of NOBr = 0.878 M, what is the equilibrium

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Answer:

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Explanation:

Given that,

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We know that,

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Initial concentration is,

$0.878\Rightarrow 0+0$

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Using formula of concentration

$k_{c}=\dfrac{[NO][Br_{2}]}{[NOBr]^2}$

Put the value into the formula

$3.07\times10^{-4}=\dfrac{[2x][x]}{[0.878-2x]^2}$

$2x^2=3.07\times10^{-4}\times(0.878)^2+3.07\times10^{-4}\times4x^2-2\times2x\times0.878\times3\times10^{-4}$

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$2x^2-0.001228x^2+0.0010536x-0.0002367=0$

$1.998772x^2+0.0010536x-0.0002367=0$

$x=0, 0.01062$

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Using formula of concentration of NO

$concentration\ of\ NO=2x$

Put the value of x

$concentration\ of\ NO=2\times0.01062$

$concentration\ of\ NO=0.02124$

Hence, The equilibrium concentration of NO is 0.02124 M.

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