Question

Помогите!!! Спасибо заранее! Решить задачу подробно!!! К 150г 20% раствора соляной кислоты добавили избыток алюминия. Рассчитайте обьем выделившегося газа!

Answer 1

2Al + 6HCl = 2AlCl3 + 3H2
m(HCl) = 150 *20/100 = 30 г
M(HCl) = 36.5 г/моль
Составим пропорцию
6 * 36.5  - 3 * 22.4
30        -      x

x=9.2 л

Answer 2

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Answer 3

Task Solution: Calculating the Volume of Gas Produced

To calculate the volume of gas produced when excess aluminum is added to 150g of 20% hydrochloric acid solution, we can use the stoichiometry of the reaction between aluminum and hydrochloric acid.

Step 1: Write the Balanced Chemical Equation

The balanced chemical equation for the reaction between aluminum and hydrochloric acid is: ``` 2 Al + 6 HCl → 2 AlCl3 + 3 H2 ```

Step 2: Calculate the Moles of Hydrochloric Acid

First, we need to calculate the moles of hydrochloric acid present in the solution.

Given: - Mass of hydrochloric acid = 150g - Concentration of hydrochloric acid = 20%

The molar mass of HCl is approximately 36.46 g/mol.

Moles of HCl = (Mass of HCl / Molar mass of HCl) * (Concentration / 100) Moles of HCl = (150 / 36.46) * (20 / 100)

Moles of HCl = 0.411 moles

Step 3: Determine the Limiting Reagent

Next, we need to determine the limiting reagent. Since aluminum is in excess, hydrochloric acid is the limiting reagent.

Step 4: Calculate the Volume of Gas Produced

Using the balanced chemical equation, we can see that 6 moles of HCl produce 3 moles of H2 gas.

Volume of gas produced = Moles of H2 * Molar volume at STP Volume of gas produced = 0.411 * 22.4 Volume of gas produced = 9.21 L

Conclusion: The volume of gas produced when excess aluminum is added to 150g of 20% hydrochloric acid solution is approximately 9.21 liters.