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kobusy [5.1K]
3 years ago
6

What symbol is used to show the standard reduction potential of an oxidation reaction in a half-cell?

Chemistry
1 answer:
Lina20 [59]3 years ago
6 0
The reduction potential of an oxidation reaction in a half-cell is denoted by the symbol E°.
Reduction potential is the tendency of a given chemical species to become reduced, that is, gain electrons or undergo a decrease in oxidation number. The greater the reduction potential, the more likely that a species will be reduced.
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A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi
Ede4ka [16]

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of Na = 5.0 g

Mass of Cl_2 = 10.0 g

Molar mass of Na = 23 g/mol

Molar mass of Cl_2 = 71 g/mol

First we have to calculate the moles of Na and Cl_2.

\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}

\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol

and,

\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}

\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

2Na+Cl_2\rightarrow 2NaCl

From the balanced reaction we conclude that

As, 2 mole of Na react with 1 mole of Cl_2

So, 0.217 moles of Na react with \frac{0.217}{2}=0.108 moles of Cl_2

From this we conclude that, Cl_2 is an excess reagent because the given moles are greater than the required moles and Na is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NaCl

From the reaction, we conclude that

As, 2 mole of Na react to give 2 mole of NaCl

So, 0.217 mole of HCl react to give 0.217 mole of NaCl

Now we have to calculate the mass of NaCl

\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl

Molar mass of NaCl = 58.5 g/mole

\text{ Mass of }NaCl=(0.217moles)\times (58.5g/mole)=12.7g

Now we have to calculate the percent yield of this reaction.

Percent yield = \frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100

Actual yield = 5.24 g

Theoretical yield = 12.7 g

Percent yield = \frac{5.24g}{12.7g}\times 100

Percent yield = 41.3 %

Therefore, the percent yield of this combination reaction is 41.3 %

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3 years ago
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Convert mass of Al to moles

Do the mole to mole ratio between Al and H2

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