Question

Помогите! Умоляю вас! Сколько моль оксида Al образуется при сгорании 2,7 г Al, если М ( Al ) = 27г / моль Пожалуйста помогите

Answer 1

Дано: m(Al) = 2,7г
Найти: n(Al)-?

Решение:
4Al + 3О2= 2Al2O3
n(Al) = m/M = 2,7г/27г/моль = 0,1 моль
По уравнению реакции:
2n(Al2O3) = 4n(Al),  

n(Al2O3) = 2n(Al)

n(Al2O3) = 1/2 = 0,05моль

Answer 2

Calculating the Number of Moles of Aluminum Oxide Formed

To calculate the number of moles of aluminum oxide formed when 2.7 g of aluminum (Al) undergoes combustion, we can use the stoichiometry of the reaction and the molar mass of aluminum.

The balanced chemical equation for the combustion of aluminum can be represented as:

2 Al + (3/2) O2 → Al2O3

First, we need to calculate the number of moles of aluminum (Al) in 2.7 g using the molar mass of aluminum.

Step 1: Calculate the number of moles of aluminum (Al) The molar mass of aluminum (Al) is 27 g/mol.

Using the formula: Number of moles = Mass / Molar mass

We can calculate: Number of moles of Al = 2.7 g / 27 g/mol Number of moles of Al = 0.1 mol

Now, we can use the stoichiometry of the balanced chemical equation to find the number of moles of aluminum oxide (Al2O3) formed.

Step 2: Use the stoichiometry of the balanced chemical equation From the balanced chemical equation: 2 moles of Al produce 1 mole of Al2O3

Using the given number of moles of Al: Number of moles of Al2O3 = 0.1 mol Al * (1 mol Al2O3 / 2 mol Al) Number of moles of Al2O3 = 0.05 mol Al2O3

Therefore, when 2.7 g of aluminum undergoes combustion, 0.05 moles of aluminum oxide (Al2O3) are formed.

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