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3 years ago

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How many moles of aluminum oxide would form if 12.5 moles of aluminum burned

1 answer:

jarptica [38.1K]3 years ago

7 0

<h3>Al + O2 -> Al2O3</h3>

Balance it:

<h3>2Al + 3O2 -> 2Al2O3</h3><h3 />

So you need 2 Al and 3 O2 to make 2 Al2O3 (aluminum oxide).

I'm going to assume you have all the O2 you need.

Since 2 mols of Al is needed to make 2 mols of the product, it's a 1:1 ratio. You get as much aluminum oxide for as much aluminum you burn.

So 12.5 mols if there is not a lack of the O2.

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What does AS> O mean?

Vlad1618 [11]

Answer:

ΔS> 0 means Letter A

Explanation:

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Carbon burns in the presence of oxygen to give carbon dioxide. Which chemical equation describes this reaction?

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3 years ago

Read 2 more answers

An unknown compound with a molar mass of 223.94 g/mol consists of 32.18% c, 4.50% h, and 63.32% cl. find the molecular formula f

Dima020 [189]

The actual number of atoms of each element present in the molecule of the compound is represented by the formula known as molecular formula.

Molar mass of the unknown compound = 223.94 g/mol (given)

Mass of each element present in the unknown compound is determined as:

Mass of carbon, $C$ :

$\frac{32.18}{100}\times 223.94 = 72.06 g$

Mass of hydrogen, $H$ :

$\frac{4.5}{100}\times 223.94 = 10.08 g$

Mass of chlorine, $Cl$ :

$\frac{63.32}{100}\times 223.94 = 141.79 g$

Now, the number of each element in the unknown compound is determined by the formula:

$number of moles = \frac{given mass}{molar mass}$

Number of moles of $C$ :

$number of moles = \frac{72.06}{12} = 6.005 mole\simeq 6 mole$

Number of moles of $H$ :

$number of moles = \frac{10.08}{1} = 10.08 mole\simeq 10 mole$

Number of moles of $Cl$

$number of moles = \frac{141.79}{35.5} = 3.99 mole\simeq 4 mole$

Dividing each mole with the smallest number of mole, to determine the empirical formula:

$C_{\frac{6}{4}}H_{\frac{10}{4}}Cl_{\frac{10}{4}}$

$C_{1.5}H_{2.5}Cl_{1}$

Multiplying with 2 to convert the numbers in formula into a whole number:

So, the empirical formula is $C_{3}H_{5}Cl_{2}$ .

Empirical mass = $12\times 3+1\times 5+2\times 35.5 = 112 g/mol$

In order to determine the molecular formula:

n = $\frac{molar mass}{empirical mass}$

n = $\frac{223.94}{112} = 1.99 \simeq 2$

So, the molecular formula is:

$2\times C_{3}H_{5}Cl_{2} = C_{6}H_{10}Cl_{4}$

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Answer:

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Explanation:

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