Question

Какой объем водорода выделится при взаимодействии серной кислоты и 5.4г алюминия

Answer 1

Дано:
m(Al)=5.4 г
Найти:
V(H2)=?
Решение:
 0.2 моль                         х моль
2Al+3H2SO4=Al2(SO4)3+3H2
2 моль                            3 моль
n(Al)=m/M=5.4 г / 27 г/моль = 0.2 моль
0.2/2=х/3
0.6=2х
х=0.3 моль (H2)
V(H2)=Vm*n=22.4 л/моль * 0.3 моль= 6.72 л
Ответ: 6.72 л

Answer 2

Reaction of Aluminum with Sulfuric Acid

When aluminum reacts with sulfuric acid, hydrogen gas is produced. The balanced chemical equation for the reaction is:

2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g)

This equation shows that for every 2 moles of aluminum, 3 moles of sulfuric acid are required, and it produces 3 moles of hydrogen gas.

Calculation of Hydrogen Gas Volume

Given that 5.4g of aluminum is used in the reaction, we can calculate the volume of hydrogen gas produced using the ideal gas law and molar volume of a gas at standard temperature and pressure (STP).

First, we need to convert the mass of aluminum to moles using its molar mass. The molar mass of aluminum is approximately 27 g/mol.

5.4g Al × (1 mol Al / 27 g Al) = 0.2 mol Al

According to the balanced chemical equation, 2 moles of aluminum produce 3 moles of hydrogen gas. Therefore, 0.2 moles of aluminum will produce:

0.2 mol Al × (3 mol H2 / 2 mol Al) = 0.3 mol H2

At STP (standard temperature and pressure), the molar volume of a gas is approximately 22.4 L/mol. Therefore, the volume of hydrogen gas produced can be calculated as:

0.3 mol H2 × 22.4 L/mol = 6.72 L

So, approximately 6.72 liters of hydrogen gas will be produced when 5.4g of aluminum reacts with sulfuric acid.