Question

1)0.672 л. кислорода прореагировало с алюминием найти массу Al2O32)Cu(OH)2 массой 9,8 г. прореагировал с HCl массой 8.3 г. определить массу соли и непрореагировавшего вещества

Answer 1

1)Дано:
V(O2)=0,672л
Найти:
m(Al2O3)=?
Решаем.
Уравнение реакции:
2Al+3O2---->2Al2O3
1) n=V:Vm, где Vm - постоянная величина равная 22,4дм3/моль
n(O2)=0,672:22,4=0,03моль
2)n(Al2O3)=(0,03*2):3=0,02моль (По уравнению реакции составил пропорцию: 0,03---------3, как x---------2).
3)M(Al2O3)=27*2+16*3=102г/моль
4)n=m:M => m=n*M => m(Al2O3)=0,02*102=2,04г
Ответ: масса равна 2,04г

Другое не успеваю, извините. :)

Answer 2

Calculation 1: Reaction between Oxygen and Aluminum

To find the mass of Al2O3 formed when 0.672 g of oxygen reacts with aluminum, we need to determine the stoichiometry of the reaction. The balanced chemical equation for the reaction between oxygen and aluminum 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 Al2O3. We can use this information to calculate the molar mass of Al2O3 and then convert the given mass of oxygen to moles.

1 mole of Al2O3 has a molar mass of 101.96 g/mol (2 Al: 2 * 26.98 g/mol + 3 O: 3 * 16.00 g/mol).

Using the molar mass of Al2O3, we can calculate the number of moles of oxygen:

0.672 g of oxygen * (1 mol O2 / 32.00 g) = 0.021 mol O2

Since the stoichiometry of the reaction is 4:3 between aluminum and oxygen, we can calculate the number of moles of aluminum:

0.021 mol O2 * (4 mol Al / 3 mol O2) = 0.028 mol Al

Now, we can calculate the mass of Al2O3 formed:

0.028 mol Al * (101.96 g Al2O3 / 2 mol Al2O3) = 1.43 g Al2O3

Therefore, when 0.672 g of oxygen reacts with aluminum, the mass of Al2O3 formed is approximately 1.43 g.