Question

ЗАДАЧА масса фосфора 3,1 г найти объем оксида азота

Answer 1

4P+5O2 = 2P2O5 - сначала пишете уравнение, уравниваете, расставляете коэффициенты

Затем нам дан объем кислорода, мы можем найти его моль по формуле: n= V/Vm, где v - объем, а Vm - молярный объем, который является 22.4 л/моль для газов при нормальных условиях:
n(O2) = 11.2 л / 22.4 (л)/моль = 0.5 моль

Теперь надо найти моли фосфора и оксида.
n(P) = 0,5 / 5 x 4 = 0.4 моль - так как перед кислородом стоит 5, 0.5 делим на 5, а так как перед фосфором стоит 4, умножаем на 4 и получаем моли фосфора. Точно так же с оксидом:
n(P2O5) = 0.5 / 5 x 2 = 0.2 моль

теперь мы можем найти их массы по формуле: m = n x M, где М - молярная масса, которую ищем в таблице Менделеева

m(P) = 0.4 моль х 31 г/моль = 12.4 г
m(P2O5) = 0.2 моль х (31х2+16х5) г/моль= 28.4 г

Answer 2

Task:

The task is to find the volume of nitrogen oxide given the mass of phosphorus as 3.1 g.

Solution:

To find the volume of nitrogen oxide, we can use the molar ratio between phosphorus and nitrogen oxide.

The balanced chemical equation for the reaction between phosphorus and nitrogen oxide is: 4P + 5NO → P4O10 + N2

From the balanced chemical equation, we can see that 4 moles of phosphorus react with 5 moles of nitrogen oxide to produce 1 mole of P4O10 and 1 mole of nitrogen gas (N2).

Given that the mass of phosphorus is 3.1 g, we can calculate the moles of phosphorus using its molar mass.

The molar mass of phosphorus (P) is approximately 30.97 g/mol.

Using the formula: moles = mass / molar mass

We can calculate the moles of phosphorus: moles of P = 3.1 g / 30.97 g/mol

Calculation:

moles of P = 3.1 g / 30.97 g/mol = 0.1 mol

Now, using the molar ratio from the balanced chemical equation, we can calculate the moles of nitrogen oxide required for the reaction.

From the balanced chemical equation: 4 moles of P react with 5 moles of NO

So, the moles of nitrogen oxide required can be calculated as: moles of NO = (moles of P) * (5/4)

Final Calculation:

moles of NO = 0.1 mol * (5/4) = 0.125 mol

Now, to find the volume of nitrogen oxide, we can use the ideal gas law equation: PV = nRT

Where: - P = pressure - V = volume - n = moles of gas - R = ideal gas constant - T = temperature

Given that the temperature and pressure are not provided, we can assume standard temperature and pressure (STP) for the calculation: - T = 273 K - P = 1 atm

At STP, 1 mole of any gas occupies 22.4 liters.

Final Answer:

Therefore, the volume of nitrogen oxide produced from 3.1 g of phosphorus at STP is approximately 0.125 mol * 22.4 L/mol = 2.8 L.