Question

Вычислить массу алюминия вступившего в реакцию с 300г раствора серной кислоты массовая доля 4,9% сказали задача на 2 действия

Answer 1

2Al + 3H2SO4 = Al2(SO4)3 + 3H2
m(H2SO4) = m(p-pa)xW(H2SO4) = 300x0,049 = 14,7 г
n(H2SO4) = m/M = 14,7/98 = 0,15 моль
Из уравнения реакции следует: n(Al) = [n(H2SO4)/3]x2 = [0,15/3]x2 = 0,1 моль
m(Al) = n(Al)xM(Al) = 0,1x27 = 2,7 г.

Answer 2

Calculation of Aluminum Mass in Reaction with Sulfuric Acid Solution

To calculate the mass of aluminum that reacted with a 300g solution of sulfuric acid with a mass fraction of 4.9%, we need to follow a few steps.

Step 1: Determine the mass of sulfuric acid in the solution. The mass fraction of sulfuric acid is given as 4.9%. This means that in 100g of the solution, there are 4.9g of sulfuric acid. Therefore, in a 300g solution, the mass of sulfuric acid can be calculated as follows:

Mass of sulfuric acid = (mass fraction of sulfuric acid / 100) * mass of the solution Mass of sulfuric acid = (4.9 / 100) * 300g

Let's calculate the mass of sulfuric acid: Mass of sulfuric acid = (4.9 / 100) * 300g = 14.7g

Step 2: Determine the molar ratio between aluminum and sulfuric acid. The balanced chemical equation for the reaction between aluminum and sulfuric acid is as follows:

2Al + 3H2SO4 → Al2(SO4)3 + 3H2

From the equation, we can see that the molar ratio between aluminum and sulfuric acid is 2:3. This means that for every 2 moles of aluminum, 3 moles of sulfuric acid are required.

Step 3: Calculate the moles of sulfuric acid. To calculate the moles of sulfuric acid, we need to use its molar mass. The molar mass of sulfuric acid (H2SO4) is approximately 98.09 g/mol.

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