Question

Сколько литров водорода выделится при реакции 6г натрия с водой?

Answer 1

Ответ:

2,9л.

Объяснение:

  6г                              Х л.

2Na+2H2O=2NaOH+H2

46                                 22,4л.

х=(6*22,4)/46=2,9 л.

Answer 2

Reaction of Sodium with Water

When 6 grams of sodium reacts with water, it produces hydrogen gas. To determine the volume of hydrogen gas produced, we can use the ideal gas law, which states that at standard temperature and pressure (STP), one mole of any gas occupies 22.4 liters.

Calculation

First, we need to calculate the number of moles of sodium reacting with water. The molar mass of sodium is 22.99 g/mol. Therefore, the number of moles of sodium is:

\[ \frac{6 \text{ g}}{22.99 \text{ g/mol}} \approx 0.261 \text{ mol} \]

According to the balanced chemical equation for the reaction of sodium with water:

\[ 2 \text{ Na} + 2 \text{ H}_2\text{O} \rightarrow 2 \text{ NaOH} + \text{H}_2 \]

We can see that 2 moles of sodium produce 1 mole of hydrogen gas. Therefore, 0.261 moles of sodium will produce:

\[ 0.261 \text{ mol} \times \frac{1 \text{ mol H}_2}{2 \text{ mol Na}} = 0.1305 \text{ mol H}_2 \]

At STP, 0.1305 moles of hydrogen gas will occupy:

\[ 0.1305 \text{ mol} \times 22.4 \text{ L/mol} = 2.924 \text{ L} \]

The volume of hydrogen gas produced when 6 grams of sodium reacts with water is approximately 2.924 liters.

Conclusion

In conclusion, when 6 grams of sodium react with water, approximately 2.924 liters of hydrogen gas are produced at standard temperature and pressure.