Question

10 г алюминия растворили в избытке серной кислоты, сколько литров водорода выделилось?

Answer 1

M(Al) = 27 г/моль
n(Al) = 10 г/27 г/моль = 0.37 моль
2Al + 3H2SO4 = Al2(SO4)3 + 3H2
Из УХР следует, что n(H2) = 1.5n(Al)
n(H2) = 1,5*0,37 моль = 0,555 моль
Vm = 22,4 л/моль
V(H2) = n(H2)*Vm
V(H2) = 0,555 моль*22,4 л/моль = 12,432 л

Answer 2

2Al+6hcl=2Alcl(ос)+3h2
n(al)= 10/27=0.37
n(h2)=0.37*3/2=0,55
V(h2)= 0,55*22.4=12.44л

Answer 3

Calculating the Volume of Hydrogen Gas Produced

To calculate the volume of hydrogen gas produced when 10g of aluminum is dissolved in excess sulfuric acid, we can use the chemical equation for the reaction between aluminum and sulfuric acid. The balanced chemical equation for the reaction is as follows:

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

According to the stoichiometry of the balanced chemical equation, 2 moles of aluminum react with 3 moles of sulfuric acid to produce 3 moles of hydrogen gas.

Calculation Steps

1. Calculate the number of moles of aluminum. 2. Use the mole ratio from the balanced chemical equation to find the number of moles of hydrogen gas produced. 3. Convert the moles of hydrogen gas to volume using the ideal gas law.

Step 1: Calculate the Number of Moles of Aluminum

The molar mass of aluminum (Al) is approximately 26.98 g/mol.

Number of moles of aluminum = Mass of aluminum / Molar mass of aluminum Number of moles of aluminum = 10g / 26.98 g/mol

Number of moles of aluminum ≈ 0.371 moles

Step 2: Use the Mole Ratio to Find the Moles of Hydrogen Gas Produced

From the balanced chemical equation, we know that 2 moles of aluminum produce 3 moles of hydrogen gas.

Number of moles of hydrogen gas = (Number of moles of aluminum) * (3 moles of H2 / 2 moles of Al) Number of moles of hydrogen gas ≈ 0.371 moles * (3/2) ≈ 0.5565 moles

Step 3: Convert Moles of Hydrogen Gas to Volume

Using the ideal gas law, we can convert the moles of hydrogen gas to volume at standard temperature and pressure (STP).

At STP, 1 mole of any gas occupies approximately 22.4 liters.

Volume of hydrogen gas = Number of moles of hydrogen gas * 22.4 L/mol Volume of hydrogen gas ≈ 0.5565 moles * 22.4 L/mol ≈ 12.47 liters

Therefore, approximately 12.47 liters of hydrogen gas would be produced when 10g of aluminum is dissolved in excess sulfuric acid