Question

Сколько граммов алюминия содержится в 51 грамме оксида алюминия

Answer 1

Mr(A2O3)=27*2+16*3=102
102 г содержит 54 г алюминия
51 грамм содержит Х г
составим пропорцию
102/51=54/x
x=51*54/102
x=27 г

Answer 2

Calculating Aluminum Content in Aluminum Oxide

To calculate the amount of aluminum in 51 grams of aluminum oxide, we can use the molar mass of aluminum oxide and the molar mass of aluminum.

Step 1: Find the Molar Mass of Aluminum Oxide The molar mass of aluminum oxide (Al2O3) can be calculated as follows: - Aluminum (Al) has a molar mass of 26.98 g/mol. - Oxygen (O) has a molar mass of 16.00 g/mol.

The molar mass of aluminum oxide (Al2O3) = (2 * molar mass of aluminum) + (3 * molar mass of oxygen) = (2 * 26.98 g/mol) + (3 * 16.00 g/mol) = 102.00 g/mol.

Step 2: Calculate the Amount of Aluminum in 51 grams of Aluminum Oxide Using the molar mass of aluminum oxide, we can calculate the amount of aluminum in 51 grams of aluminum oxide: - Number of moles of aluminum oxide = mass / molar mass = 51 g / 102.00 g/mol = 0.5 moles. - Since there are 2 moles of aluminum in 1 mole of aluminum oxide, the amount of aluminum in 51 grams of aluminum oxide is 2 * 0.5 moles = 1 mole.

Therefore, there are approximately 26.98 grams of aluminum in 51 grams of aluminum oxide.

Conclusion Approximately 26.98 grams of aluminum are contained in 51 grams of aluminum oxide.