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3 years ago

2A1 (s) + 3C12 (g) --> 2AlCl3 (s) (balanced)

Chemistry

1 answer:

Dennis_Churaev [7]3 years ago

5 0

Answer:

32.3%

Explanation:

Percent yield is defined as:

Actual yield (125.5g) / Theoretical Yield * 100

To find theoretical yield we have to find the moles of aluminium. As 2 moles of Al produce 2 moles of AlCl3, the moles of Al = Moles AlCl3.

With these moles we can find the mass assuming a 100% of yield (Theoretical Yield) as follows:

<em>Moles Al = Moles AlCl3 (Molar mass Al = 26.98g/mol)</em>

72g Al * (1mol / 26.98g) = 2.67 moles AlCl3

<em>Mass AlCl3 (Molar mass: 133.34g/mol)</em>

2.67 moles AlCl3 * (133.34g / mol) = 355.8g AlCl3

Percent Yield = 125.5g / 355.8g * 100 =

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TEA [102]

Explanation:

The scientific notation:

$a\times10^k$

where

$1\leq a$ and k is integer.

We have the example:

$(5\times10^3)+(4.3\times10^4)$

You can write the numbers in a "normal" form:

$5\times10^3=5\times1000=5000\\\\4.3\times10^4=4.3\times10000=43000$

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And next write it in the scientific notation:

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<h3>Other method:</h3>

You can add numbers in scientific notation if the power of tens in both number is the same.

Therefore you must convert the first number:

$5\times10^3=0.5\times10\times10^3=0.5\times10^4$

Now, you can make the sum:

$(5\times10^3)+(4.3\times10^4)=0.5\times10^4+4.3\times10^4=(0.5+4.3)\times10^4=4.8\times10^4$

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3 years ago

What would the products be when aluminum chloride (which contains Al3+ and Cl- ions) is melted and electrolyzed? Write half equa

Hoochie [10]

The product of electrolysis : Al at cathode and Cl₂(chlorine) at anode

<h3>Further explanation</h3>

Given

Aluminum chloride compound

Required

The product of electrolysis

Solution

The rule :

The reaction at the cathode(the negative pole) :

1. the reduced active metal is water, other than that the metal will be reduced

2. H⁺ of the acid will be reduced

The reaction at the anode((the positive pole) :

1. if the electrodes are not inert then the metal is oxidized

2. If inert then:

a. OH⁻ from the base will be oxidized

b. The halogen metal will oxidize

In the electrolysis of molten AlCl₃ with an inert electrode, the cation will be reduced at the cathode and the anion will be oxidized at the anode

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Anode : 2Cl⁻⇒Cl₂(g) + 2e⁻

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Ainat [17]

Answer:

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Explanation:

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AlladinOne [14]

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