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3 years ago

How many moles of CaCO3 are needed to react with 12.5 mol SO2

Chemistry

1 answer:

grandymaker [24]3 years ago

7 0

<h3><u>Answer;</u></h3>

= 12.5 Moles of CaSO3

<h3><u>Explanation</u>;</h3>

The reaction between CaCO3 and SO2 is given by the equation.

CaCO3(s) + SO2(g) → CaSO3(aq) + CO2(g)

The mole ratio between CaCO3 and SO2 is 1 : 1;

1 mole of CaCO3 reacts with 1 mole SO2 to form CaSO3 and CO2

Therefore;

<em>12.5 moles of SO2 will require 12.5 moles of CaSO3</em>

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Calculate the electrical energy per gram of anode material for the following reaction at 298 K:

solmaris [256]

43.8 kJ

<h3>Explanation</h3>

There are two electrodes in a voltaic cell. Which one is the anode?

The lithium atom used to have no oxygen atoms when it was on the reactant side. It gains two oxygen atoms after the reaction. It has gained more oxygen atoms than the manganese atom. Gaining oxygen is oxidation. As a result, lithium is being oxidized.

Oxidation takes place at the anode of a cell. Therefore, the anode of this cell is made of lithium.

Lithium has an atomic mass of 6.94. Each gram of Li would contain 1/6.94 = 0.144 moles of Li atoms. Each Li atom loses one electron in this cell. Therefore, the number of electron transferred, <em>n</em>, equals 0.144 moles for each gram of the anode.

Let $w_\text{max}$ represents the electrical energy produced.

$w_\text{max} = n \cdot F \cdot E_\text{cell}$ , where

<em>n</em> is the <em>number of moles</em> electrons transferred,
<em>F</em> is the Faraday's constant,
<em>E</em> $_\text{cell}$ is the cell potential,

<em>n </em>= 0.144 mol, as shown above, and

<em>F </em>= 96.486 kJ / ( $\text{V} \cdot \text{mol} \; \text{e}^{-}$ ).

Therefore,

$w_\text{max} = n \cdot F \cdot E\\\phantom{w_\text{max}} = 0.144 \times 96.486 \times 3.15 \\\phantom{w_\text{max}} = 43.8 \; \text{kJ}$ .

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The explanation:

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