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nalin [4]
3 years ago
12

The nahco3 is the limiting reactant and the hcl is the excess reactant in this experiment. determine the theoretical yield of th

e nacl product, showing all of your work in the space below.
Chemistry
1 answer:
galben [10]3 years ago
4 0
<span>37.06g - 24.35g (evaporating dish) = 12.71 g NaHCO3


since
1 mol NaHCO3 & 1 HCl --> 1 mol NaCl & 1 H2O & 1 CO2
use molar masses to find mass of NaCl produced:

12.71 g NaHCO3 @ 58.44 g/mol NaCl / 84.01 g/mol NaHCO3 = 8.841 grams of NaCl </span>
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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

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

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

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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}.

3 0
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