Question

ПОМОГИТЕ!!! №1Вычислите объем кислорода (н.у.), необходимый для сжигания 2,7г аллюминия.( Такая реакция происходит при поджигании бенгальских огней.)№2Вычислите объемы водорода и кислорода (н.у.), которые образуются при разложении 90г воды.

Answer 1

1.

Пишем уравнение реакции:

 2,7г              хл

 4Al      +       3O2===>2Al2O3

M=27г/моль  Vm=22,4л/моль

m=108 г        V=67,2 л

Пропорция: 2,7/108=x/67,2

Выводим: x=2,7*67,2/108=1.68 л.

Ответ:1,68 л

2.

 Уравнение:

 90 г         х л         х л

2H2O===>2H2    +   O2 

Пусть: 

1 моль     1 моль    1/2 моль

v(H2O)=m/M=90г/18г/моль=5 моль

v(H2) =5 моль

v(O2)=2,5 моль

V(H2)=Vm*v=5 моль*22,4 л/моль=112 л

V(O2)=Vm*v=2,5 моль* 22,4 л/моль= 56 л

Ответ:112 л;56 л. 

Answer 2

Answer:

№1 To calculate the volume of oxygen (at standard conditions) required to burn 2.7 grams of aluminum, we need to use the balanced chemical equation for the reaction between aluminum and oxygen. The balanced equation is:

4 Al + 3 O2 → 2 Al2O3

From the balanced equation, we can see that 4 moles of aluminum react with 3 moles of oxygen to produce 2 moles of aluminum oxide.

To find the moles of aluminum, we divide the given mass by the molar mass of aluminum:

Molar mass of aluminum (Al) = 26.98 g/mol

Moles of aluminum = mass of aluminum / molar mass of aluminum Moles of aluminum = 2.7 g / 26.98 g/mol

Now, using the mole ratio from the balanced equation, we can calculate the moles of oxygen required:

Moles of oxygen = (moles of aluminum / 4) * 3

Finally, we can convert the moles of oxygen to volume using the ideal gas law:

Volume of oxygen = moles of oxygen * molar volume of gas at standard conditions

The molar volume of gas at standard conditions (0 degrees Celsius and 1 atmosphere pressure) is approximately 22.4 liters/mol.

Let's calculate the volume of oxygen:

Molar mass of oxygen (O2) = 32 g/mol

Moles of aluminum = 2.7 g / 26.98 g/mol = 0.1 mol Moles of oxygen = (0.1 mol / 4) * 3 = 0.075 mol Volume of oxygen = 0.075 mol * 22.4 L/mol

Therefore, the volume of oxygen required to burn 2.7 grams of aluminum is approximately 1.68 liters.

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