Question

ДАЮ 35 БАЛЛОВ ЗА ЗАДАЧУ !!! УЧУСЬ В 9 КЛАССЕ. Образец технического алюминия m=50 грамм (4 % примеси) обработали соляной кислотой.Каков объем полученного водорода?

Answer 1

2Al+6HCl=2AlCl3+3H2 вот уравнение теперь у нас 50 г алюминия с примесями 4% значит чистого алюминия =50*0.96=48 г использовался в реакции теперь моль =48/27=1.78 моль соотношение 2:3 значит 1.78/2*3=2.67 моль а объем равен 2.67*22.4=59.8 л

Answer 2

Calculation of the Volume of Hydrogen Gas Produced

To calculate the volume of hydrogen gas produced when treating a sample of technical aluminum with hydrochloric acid, we need to know the molar mass of aluminum and the balanced chemical equation for the reaction.

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

The balanced chemical equation for the reaction between aluminum and hydrochloric acid (HCl) is as follows:

2 Al + 6 HCl -> 2 AlCl3 + 3 H2

From the balanced equation, we can see that 2 moles of aluminum react with 6 moles of hydrochloric acid to produce 3 moles of hydrogen gas.

Given that the mass of the aluminum sample is 50 grams and it contains 4% impurities, we can calculate the mass of pure aluminum in the sample:

Mass of pure aluminum = 50 grams - (4% of 50 grams) = 50 grams - (0.04 * 50 grams) = 50 grams - 2 grams = 48 grams

Now, we can calculate the number of moles of aluminum in the sample:

Number of moles of aluminum = Mass of pure aluminum / Molar mass of aluminum Number of moles of aluminum = 48 grams / 26.98 g/mol ≈ 1.78 moles

According to the balanced equation, 2 moles of aluminum react to produce 3 moles of hydrogen gas. Therefore, the number of moles of hydrogen gas produced can be calculated as follows:

Number of moles of hydrogen gas = (3 moles of H2 / 2 moles of Al) * Number of moles of aluminum Number of moles of hydrogen gas = (3/2) * 1.78 moles ≈ 2.67 moles

Finally, we can use the ideal gas law to calculate the volume of hydrogen gas produced:

PV = nRT

Where: P = pressure (assumed to be constant) V = volume of gas n = number of moles of gas R = ideal gas constant T = temperature (assumed to be constant)

Since the pressure and temperature are assumed to be constant, we can rearrange the equation to solve for the volume:

V = (n * R * T) / P

Assuming standard temperature and pressure (STP), which is 273.15 K and 1 atm respectively, and using the ideal gas constant (R) value of 0.0821 L·atm/(mol·K), we can calculate the volume of hydrogen gas produced:

V = (2.67 moles * 0.0821 L·atm/(mol·K) * 273.15 K) / 1 atm V ≈ 58.5 liters

Therefore, the volume of hydrogen gas produced when treating the 50-gram sample of technical aluminum with hydrochloric acid is approximately 58.5 liters.

Please note that this calculation assumes ideal conditions and may not account for all factors that could affect the actual volume of hydrogen gas produced.