Question

Масса оксида алюминия образовавшегося при горении алюминия с 16 гр кислорода равно 1) 17 гр. 2)34 гр. 3) 68 гр

Answer 1

16 г. Х г
4Al. +. 3O2. =. 2Al2O3
n=3 моль. n=2 моль
М=32 г/моль. М=102 г/моль
m=96 г. m=204 г
16 г О2. -. Х г Al2O3
96 г О2. -. 204 г Al2O3
m(Al2O3)=16*204/96=34 г
Ответ:2)

Answer 2

To find the mass of aluminum oxide formed during the combustion of aluminum with 16 grams of oxygen, we need to understand the chemical reaction involved.

The chemical equation for the combustion of aluminum with oxygen is as follows:

4 Al + 3 O2 -> 2 Al2O3

This equation shows that 4 moles of aluminum react with 3 moles of oxygen to produce 2 moles of aluminum oxide.

The molar mass of aluminum (Al) is approximately 27 g/mol, and the molar mass of oxygen (O2) is approximately 32 g/mol (16 g/mol per oxygen atom).

Now, let's calculate the moles of oxygen present:

Number of moles of oxygen = Mass of oxygen / Molar mass of oxygen Number of moles of oxygen = 16 g / 32 g/mol = 0.5 moles

According to the balanced chemical equation, 4 moles of aluminum react with 3 moles of oxygen to produce 2 moles of aluminum oxide. Since the molar ratio is 4:3, we can find the number of moles of aluminum oxide formed:

Number of moles of aluminum oxide = (Number of moles of oxygen / 3) * 2 Number of moles of aluminum oxide = (0.5 moles / 3) * 2 = 0.333 moles

Finally, let's calculate the mass of aluminum oxide formed:

Mass of aluminum oxide = Number of moles of aluminum oxide * Molar mass of aluminum oxide Mass of aluminum oxide = 0.333 moles * (2 * molar mass of aluminum) (since Al2O3 has two aluminum atoms) Mass of aluminum oxide = 0.333 moles * (2 * 27 g/mol) = 18 g

So, the mass of aluminum oxide formed during the combustion of aluminum with 16 grams of oxygen is approximately 18 grams.

None of the provided options (17 g, 34 g, 68 g) matches the calculated value. The correct answer would be approximately 18 g.