Question

Рассчитать массу и объем кислорода необходимого для взаимодействия с 6,2 г фосфора в результате реакции образуеться оксид фосфора 5

Answer 1

Напишем уравнения реакции:
4P+ 5O2 = 2P2O5;
Поскольку масса Фосфора = 6,2 складываем пропорцию
$\frac{4}{6,2} = \frac{5}{x}$
4х = 31
х = 7,75
Чтобы вступать в реакцию 6,2 г фосфора нужно 7,75 г кислорода. 

Answer 2

Calculating the Mass and Volume of Oxygen Required for the Reaction

To calculate the mass and volume of oxygen required for the reaction with 6.2 g of phosphorus to form phosphorus oxide (P2O5), we can use stoichiometry to determine the amount of oxygen needed.

Step 1: Determine the Balanced Chemical Equation The balanced chemical equation for the reaction of phosphorus with oxygen to form phosphorus oxide (P2O5) is: ``` 4P + 5O2 → P2O5 ``` This equation shows that 4 moles of phosphorus react with 5 moles of oxygen to produce 2 moles of phosphorus oxide.

Step 2: Calculate the Moles of Phosphorus Given mass of phosphorus = 6.2 g Molar mass of phosphorus (P) = 30.97 g/mol

Moles of phosphorus = Given mass / Molar mass Moles of phosphorus = 6.2 g / 30.97 g/mol Moles of phosphorus ≈ 0.20 mol

Step 3: Calculate the Moles of Oxygen Using the stoichiometry of the balanced chemical equation, we can determine the moles of oxygen required.

From the balanced equation, 4 moles of phosphorus react with 5 moles of oxygen. So, moles of oxygen = (moles of phosphorus) * (5 moles of oxygen / 4 moles of phosphorus) Moles of oxygen ≈ 0.25 mol

Step 4: Calculate the Mass of Oxygen Molar mass of oxygen (O2) = 32 g/mol

Mass of oxygen = Moles of oxygen * Molar mass of oxygen Mass of oxygen ≈ 0.25 mol * 32 g/mol Mass of oxygen ≈ 8 g

Step 5: Calculate the Volume of Oxygen at STP At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters.

Volume of oxygen at STP = Moles of oxygen * 22.4 L/mol Volume of oxygen ≈ 0.25 mol * 22.4 L/mol Volume of oxygen ≈ 5.6 L

Therefore, the mass of oxygen required for the reaction with 6.2 g of phosphorus to form phosphorus oxide (P2O5) is approximately 8 grams, and the volume of oxygen at STP is approximately 5.6 liters.