Question

Скільки грамів алюміній нітрату утвориться при взаємодії 5,1г алюміній оксиду з нітратною кислотою? ребят,пожалуйста помогите. очень нужно)

Answer 1

Объяснение:

5.1г. хг

Al2O3 + 6HNO3 = 2Al(NO3)3 + 3H2O

102г/моль 2•213г/моль

х=(5.1•2•213)/102=21.3г алюміній нітрату

Answer 2

Calculation of Aluminum Nitrate Formation

To calculate the amount of aluminum nitrate formed when 5.1g of aluminum oxide reacts with nitric acid, we can use the balanced chemical equation for the reaction between aluminum oxide and nitric acid.

The balanced chemical equation for the reaction is: 3 Al2O3 + 6 HNO3 → 3 H2O + 6 NO2 + 2 Al(NO3)3

From the balanced chemical equation, we can see that 3 moles of aluminum oxide react with 6 moles of nitric acid to produce 2 moles of aluminum nitrate.

Calculation Steps

1. Convert grams of aluminum oxide to moles: - The molar mass of aluminum oxide (Al2O3) is approximately 101.96 g/mol. - Number of moles of aluminum oxide = Mass of aluminum oxide / Molar mass of aluminum oxide - Number of moles of aluminum oxide = 5.1g / 101.96 g/mol

2. Use mole ratio from the balanced equation: - According to the balanced equation, 3 moles of aluminum oxide produce 2 moles of aluminum nitrate.

3. Calculate moles of aluminum nitrate: - Number of moles of aluminum nitrate = (Number of moles of aluminum oxide * 2) / 3

4. Convert moles of aluminum nitrate to grams: - Mass of aluminum nitrate = Number of moles of aluminum nitrate * Molar mass of aluminum nitrate

Let's calculate the amount of aluminum nitrate formed using the given information.