Question

Какой объем водорода может быть получен при растворении в едком натре 270 г сплава алюминия, содержащего 20% меди.

Answer 1

216мг/моль
x мл
2Al + 2NaOH+6H2O = 2Na[Al
(OH)4]+ 3H2
54мг/моль
22,4 мл/моль*
3моль=67,2мл
m(чист.Al) = 270*(1-0.2)=216 мг
Составим пропорцию, по которой х=
(216*67,2)/54 мл=268,8 мл
V(практ. Н2) = 0,85 * 268,8=228,5 мл
Ответ: 228,5 мл

Answer 2

Calculating the Volume of Hydrogen Obtained from Dissolving Aluminum-Copper Alloy in Liquid Sodium

To calculate the volume of hydrogen obtained when dissolving 270g of an aluminum-copper alloy containing 20% copper in liquid sodium, we can use the following steps:

1. Calculate the moles of aluminum and copper in the alloy. 2. Determine the moles of hydrogen produced from the reaction. 3. Convert the moles of hydrogen to volume at standard conditions.

Step 1: Calculate the Moles of Aluminum and Copper in the Alloy

The first step is to calculate the moles of aluminum and copper in the alloy. We can use the given mass of the alloy and the percentage of copper to determine the masses of aluminum and copper in the alloy.

The mass of aluminum in the alloy: - Mass of aluminum = Total mass of alloy * (1 - Percentage of copper) - Mass of aluminum = 270g * (1 - 0.20) - Mass of aluminum = 270g * 0.80 - Mass of aluminum = 216g

The mass of copper in the alloy: - Mass of copper = Total mass of alloy * Percentage of copper - Mass of copper = 270g * 0.20 - Mass of copper = 54g

Step 2: Determine the Moles of Hydrogen Produced from the Reaction

The reaction between aluminum-copper alloy and liquid sodium produces hydrogen gas. The balanced chemical equation for the reaction is needed to determine the moles of hydrogen produced.

Step 3: Convert the Moles of Hydrogen to Volume at Standard Conditions

Once the moles of hydrogen produced are determined, they can be converted to volume at standard conditions using the ideal gas law.

I will now proceed to calculate the moles of aluminum and copper in the alloy and determine the moles of hydrogen produced from the reaction.