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2 years ago

if there are 19 grams of oxygen in a sample of aluminum oxide . how many grams of aluminium are present

Chemistry

2 answers:

Maksim231197 [3]2 years ago

3 0

Explanation:

Al2O3

16=molar mass of one oxygen atom

Number of moles: mass/molar mass

Mol of O3: 19÷(16*3) = 0.40 g/mol

mole ratio

0.40: 3 (oxygen)

0.27: 2 (aluminum)

Mass = molar mass*mole

0.27*(27*3)= 21.87g

Whitepunk [10]2 years ago

3 0

Molecular Formula of (Aluminum Oxide) is Al2O3

Since the Molar Mass of Oxygen Atom = 16

Number of Moles (Atom) = Mass (Atom) / Molar Mass (Atom)

So, Number of Moles (Oxygen O3 Atom) = 19 / (16 * 3)
= 19 / 48
= 0.40 gm/mole

Molar Ratio of (Oxygen O3 Atom) = 0.40:3
Molar Ratio of (Aluminum Al2 Atom) = 0.27:2

Since, Number of Moles (Atom) = Mass (Atom) / Molecular Mass (Atom)

So, Mass (Atom) = Molar Mass (Atom) * Number of Moles (Atom)
Mass (Aluminum Al2 Atom) = 0.27 * (27 * 3)
= 0.27 * 81
= 21.87 gm

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Answer : The enthalpy of the reaction = -1839.6 KJ

Solution : Given,

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The balanced chemical reaction is,

$3MnO_{2}(s)+4Al(s)\rightarrow 2Al_{2}O_{3}(s)+3Mn(s)$

Formula used :

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$\Delta (H_{f})_{reaction}=(2\times \Delta H_{Al_{2}O_{3}(s)}+3\times \Delta H_{Mn(s)} )-(3\times \Delta H_{MnO_{2}(s) }+4\times\Delta H_{Al}(s))$

We know that the standard enthalpy of formation of the element is equal to Zero.

Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.

Now, put all the values in above formula, we get

$\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]$

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2 years ago

if there are 19 grams of oxygen in a sample of aluminum oxide . how many grams of aluminium are present ​

if there are 19 grams of oxygen in a sample of aluminum oxide . how many grams of aluminium are present