Question

Какая масса натрия необходима для взаимодействия с 44,8 литрами кислорода

Answer 1

M(Na)-?
V(02)=44.8л
Решение
Na+O2=Na2O
V(количества вещества)=44.8:22.4=2моль
m(Na)=M(Na)*V(Na)=23*2=46 кг или г, я не помню

Answer 2

Calculating the Mass of Sodium Required for Reacting with 44.8 Liters of Oxygen

To calculate the mass of sodium required for reacting with 44.8 liters of oxygen, we can use the balanced chemical equation for the reaction between sodium and oxygen. The balanced chemical equation for the reaction is:

2Na + O2 → 2Na2O

From the balanced chemical equation, we can see that 2 moles of sodium react with 1 mole of oxygen to produce 2 moles of sodium oxide. Using the molar volume of a gas at standard temperature and pressure (STP), which is 22.4 liters per mole, we can calculate the amount of sodium required.

Calculation

Given: Volume of oxygen (O2) = 44.8 liters

Using the molar volume of a gas at STP: 1 mole of O2 = 22.4 liters

Therefore, the number of moles of O2 = Volume of O2 / Molar volume of O2 = 44.8 / 22.4 = 2 moles

According to the balanced chemical equation: 2 moles of Na react with 1 mole of O2

So, the number of moles of Na required = 2 * 2 = 4 moles

Now, to find the mass of sodium required, we can use the molar mass of sodium, which is approximately 23 grams per mole.

Result

The mass of sodium required for reacting with 44.8 liters of oxygen is: 4 moles * 23 grams/mole = 92 grams.