Question

Алюминий массой 13,5 г поместили в раствор, в котором содержится серная кислота массой 158,8 г.Какой объем газа (н.у.) выделился?

Answer 1

2Al+6H2SO4=Al2(SO4)3+2SO2+6H2O
находим что в избытке,что в недостатке
m(Al)дано 13,5г
М(АL)=54
13,4/55=0,25
m(H2SO4)=158,8г
M(H2SO4)=588
158,8/588=0,27
Алюминий в недостатке,по нему ведём расчёт
Обозначим объём газа SO2 х.
Vm=22,4*3=67,2
х=(13,5*67,2)/54=16,8
ответ:16,8л

Answer 2

Calculation of Gas Volume Released

To calculate the volume of gas released when 13.5 g of aluminum is placed in a solution containing 158.8 g of sulfuric acid, we need to consider the balanced chemical equation for the reaction between aluminum and sulfuric acid.

The balanced chemical equation for the reaction is as follows: 2Al + 3H2SO4 -> Al2(SO4)3 + 3H2

From the balanced equation, we can see that 2 moles of aluminum react with 3 moles of sulfuric acid to produce 1 mole of aluminum sulfate and 3 moles of hydrogen gas.

To find the volume of gas released, we need to convert the mass of aluminum to moles and then use the mole ratio to determine the moles of hydrogen gas produced. Finally, we can use the ideal gas law to calculate the volume of gas at standard temperature and pressure (STP).

Let's calculate the volume of gas released step by step:

1. Calculate the moles of aluminum: - The molar mass of aluminum (Al) is 26.98 g/mol. - Moles of aluminum = mass of aluminum / molar mass of aluminum - Moles of aluminum = 13.5 g / 26.98 g/mol

2. Calculate the moles of sulfuric acid: - The molar mass of sulfuric acid (H2SO4) is 98.09 g/mol. - Moles of sulfuric acid = mass of sulfuric acid / molar mass of sulfuric acid - Moles of sulfuric acid = 158.8 g / 98.09 g/mol

3. Determine the limiting reactant: - To determine the limiting reactant, we compare the moles of aluminum and sulfuric acid using the stoichiometric ratio from the balanced equation. - The stoichiometric ratio between aluminum and sulfuric acid is 2:3. - Divide the moles of aluminum by 2 and the moles of sulfuric acid by 3. - The smaller value obtained will be the limiting reactant.

4. Calculate the moles of hydrogen gas produced: - The stoichiometric ratio between aluminum and hydrogen gas is 2:3. - Multiply the moles of the limiting reactant by the stoichiometric ratio to find the moles of hydrogen gas produced.

5. Calculate the volume of gas at STP: - The molar volume of gas at STP is 22.4 L/mol. - Volume of gas = moles of gas * molar volume of gas at STP

Now, let's perform the calculations:

1. Moles of aluminum: - Moles of aluminum = 13.5 g / 26.98 g/mol = 0.500 mol 2. Moles of sulfuric acid: - Moles of sulfuric acid = 158.8 g / 98.09 g/mol = 1.618 mol 3. Determining the limiting reactant: - The stoichiometric ratio between aluminum and sulfuric acid is 2:3. - Moles of aluminum / 2 = 0.500 mol / 2 = 0.250 mol - Moles of sulfuric acid / 3 = 1.618 mol / 3 = 0.539 mol - The limiting reactant is aluminum, as it produces fewer moles of product.

4. Moles of hydrogen gas produced: - The stoichiometric ratio between aluminum and hydrogen gas is 2:3. - Moles of hydrogen gas = Moles of limiting reactant * (3/2) - Moles of hydrogen gas = 0.250 mol * (3/2) = 0.375 mol

5. Volume of gas at STP: - Volume of gas = Moles of gas * Molar volume of gas at STP - Volume of gas = 0.375 mol * 22.4 L/mol = 8.4 L

Therefore, when 13.5 g of aluminum is placed in a solution containing 158.8 g of sulfuric acid, approximately 8.4 liters of gas (at standard temperature and pressure) will be released.