Question

Задача.Расчитайте объем водорода,который выделится при взаимодействии 0.23 граммов натрия с достаточным количеством воды

Answer 1

H2+2NaOH=2H2O+2Na
Берем водород за x
При н.у. Объем=22,4 л
Значит , у натрия 44,8 л(2 моль), у водорода 22,4(1 моль) . Решаем пропорцию x/22,4=0,23/44,8 ; x=0,2 л

Answer 2

Calculation of Hydrogen Gas Volume

To calculate the volume of hydrogen gas that will be produced when 0.23 grams of sodium reacts with water, we need to consider the balanced chemical equation for the reaction:

2Na + 2H2O -> 2NaOH + H2

From the equation, we can see that 2 moles of sodium (Na) react with 2 moles of water (H2O) to produce 1 mole of hydrogen gas (H2).

To calculate the moles of sodium (Na) in 0.23 grams, we need to use the molar mass of sodium, which is 22.99 g/mol.

Moles of Na = Mass of Na / Molar mass of Na Moles of Na = 0.23 g / 22.99 g/mol

Now, we can use the mole ratio from the balanced equation to calculate the moles of hydrogen gas (H2) produced.

Moles of H2 = Moles of Na / 2

Finally, we can use the ideal gas law to calculate the volume of hydrogen gas at standard temperature and pressure (STP), which is 0 degrees Celsius (273.15 Kelvin) and 1 atmosphere (101.325 kilopascals or 760 mmHg).

Volume of H2 = Moles of H2 * 22.4 L/mol

Let's calculate the volume of hydrogen gas:

Molar mass of Na = 22.99 g/mol Moles of Na = 0.23 g / 22.99 g/mol = 0.01 mol Moles of H2 = 0.01 mol / 2 = 0.005 mol Volume of H2 = 0.005 mol * 22.4 L/mol = 0.112 L

Therefore, approximately 0.112 liters of hydrogen gas will be produced when 0.23 grams of sodium reacts with water.