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2 years ago

How many grams of FeCo3 will be produced from 57.2g FeCl2

Chemistry

1 answer:

Evgesh-ka [11]2 years ago

4 0

Answer:

287.30 g of FeCO₃

Solution:

The Balance Chemical Equation is as follow,

FeCl₂ + Na₂CO₃ → FeCO₃ + 2 NaCl

Step 1: Calculate Mass of FeCl₂ as,

Molarity = Moles ÷ Volume

Solving for Moles,

Moles = Molarity × Volume

Putting Values,

Moles = 2 mol.L⁻¹ × 1.24 L

Moles = 2.48 mol

Also,

Moles = Mass ÷ M.Mass

Solving for Mass,

Mass = Moles × M.Mass

Putting Values,

Mass = 2.48 mol × 126.75 g.mol⁻¹

Mass = 314.34 g of FeCl₂

Step 2: Calculate Mass of FeCO₃ formed as,

According to equation,

126.75 g (1 mole) FeCl₂ produces = 115.85 g (1 mole) FeCO₃

So,

314.34 g of FeCl₂ will produce = X g of FeCO₃

Solving for X,

X = (314.34 g × 115.85 g) ÷ 126.75 g

X = 287.30 g of FeCO₃

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Answer:

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Explanation:

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$s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }$

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Where:

$m$ - Mass, in kilograms.

$c_{w}$ - Specific heat of water, in joules per kilogram-Kelvin.

$T_{1}$ , $T_{2}$ - Initial and final temperatures of water, in Kelvin.

If we know that $m = 1\,kg$ , $c_{w} = 4190\,\frac{J}{kg\cdot K}$ , $T_{1} = 373.15\,K$ and $T_{2} = 288.15\,K$ , then the change in entropy for the entire process is:

$s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}$

$s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}$

The change in entropy is -1083.112 joules per kilogram-Kelvin.

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