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Harrizon [31]
3 years ago
9

What is the mass of 2.56 × 10–4 moles of Fe2O3? : 6.23 × 105 159.6 g 1.60 × 0–6 g 4.09 × 10–2 g

Chemistry
1 answer:
Monica [59]3 years ago
7 0

Answer:

4.09 × 10⁻² g

Solution:

As we know that,

1 mol of Fe₂O₃ is equal to = 159.69 g

So,

2.56 × 10⁻⁴ mol will be equal to = X g

Solving for X,

X = (159.69 g × 2.56 × 10⁻⁴ mol) ÷ 1 mol

X = 0.04088 g

Or,

X = 0.049 g

Or,

X = 4.09 × 10⁻² g

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I_{2}  (g) + Cl_{2}  (g) ----->  2 ICl (g)

Step 2 : Set up ICE table

We will set up an ICE table for the above reaction

Following points are considered while drawing ICE table

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- The change in concentration (C) is assumed as x. Change (x) is negative for reactants and positive for products

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Check attached file for ICE table

Step 3 : Set up equilibrium constant equation

The equation for equilibrium constant can be written as

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Let us plug in the equilibrium values (E) for I₂, Cl₂ and ICl from ICE table

81.9 = \frac{(2x)^{2}}{(0.437-x) (0.269-x)}

81.9 = \frac{(2x)^{2}}{x^{2} -0.706x + 0.118}

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4x^{2} = 81.9x^{2} -57.8x +9.66

77.9x^{2} -57.8x+9.66 = 0

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Hence, x = 0.254 M

From the ICE table, we know that the equilibrium concentration of ICl is 2x

[ICl]_{eq} = 2 ( 0.254) = 0.508 M

The concentration of ICl when the reaction reaches equilibrium is 0.508 M

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