Question

Пожалуйста помогите , задачка по химии:Сколько литров водорода получится при взаимодействии 12 г натрия с водой, если в составе натрия 5% примесей?

Answer 1

2Na+2H2O=2NaOH+H2
1)100%-5%=95% чистого вещества.
Mr(Na)=2*23=46г/моль.
2)12*0,95=11,4(г) чистое вещество
Mr(H2)=22,4 л/моль.
Составляем пропорцию. Получаем.
11,4/46=x/22,4
x=5,6 л.
Ответ 5,6 л.

Answer 2

Task:

You need to find out how many liters of hydrogen will be produced when 12 grams of sodium react with water, given that the sodium contains 5% impurities.

Solution:

To solve this problem, we can use the chemical equation for the reaction between sodium and water. The balanced chemical equation for the reaction is:

2Na + 2H2O -> 2NaOH + H2

From the equation, we can see that 2 moles of sodium react with 2 moles of water to produce 1 mole of hydrogen gas. Now, let's calculate the amount of hydrogen gas produced.

First, we need to find the number of moles of sodium in 12 grams of sodium, taking into account the 5% impurities. Then, we can use the stoichiometry of the reaction to find the amount of hydrogen gas produced.

Calculation:

1. Calculate the moles of pure sodium: - Mass of pure sodium = 12g * (1 - 5%) = 12g * 0.95 = 11.4g - Moles of pure sodium = 11.4g / molar mass of sodium

2. Use stoichiometry to find the moles of hydrogen gas produced: - Moles of hydrogen gas = Moles of pure sodium / 2

3. Convert moles of hydrogen gas to liters at standard temperature and pressure (STP): - 1 mole of any gas at STP occupies 22.4 liters

Result:

The amount of hydrogen gas produced can be calculated using the above steps. Let's proceed with the calculations and find the final result.