Question

ПОМОГИТЕ СРОЧНО!!! ХИМИЯ 8 КЛАСС. ДАЮ 100 БАЛЛОВ 1) Найти объём кислорода для взаимодействия с железом, для получения 32г оксида железа (3). 2) Найти массу корбоната кальция при разложении, которого образуется 4,48л оксида углерода (4) и оксида кальция. ПОЖАЛУЙСТА, ПОМОГИТЕ ХОТЯ БЫ С ОДНОЙ ЗАДАЧЕЙ.

Answer 1

1)

4 Fe + 3 O2 = 2 Fe2O3

M(Fe) = 56 г/моль

n = m ÷ M

n(Fe) = m(Fe) ÷ M(Fe) = 32 г ÷ 56 г/моль = 0.57 моль

n(O2) = ¾n(Fe) = (0,57 × 3) ÷ 4 = 0.43 моль

V = Vn × n

V(O2) = Vn × n(O2) = 22,4 л/моль × 0.43 моль = 9.63 л

Ответ: V (O2) = 9.63 л

2)

CaCO3 = CaO + CO2

V = Vn × n

n(CO2) = V(CO2) ÷ Vn = 4.48 л ÷ 22,4 л/моль = 0.2 моль

n(CaCO3) = n(CO2) = 0.2 моль

M(CaCO3) = 100 г/моль

m = n × M

m(CaCO3) = n(CaCO3) × M(CaCO3) = 0.2 моль × 100 г/моль = 20 г

Ответ: m (CaCO3) = 20 г

Answer 2

Problem 1: Finding the Volume of Oxygen for the Reaction with Iron

To find the volume of oxygen required for the reaction with iron to produce 32g of iron(III) oxide (Fe2O3), we need to use the stoichiometry of the reaction.

The balanced equation for the reaction is:

4Fe + 3O2 → 2Fe2O3

From the equation, we can see that 3 moles of oxygen react with 4 moles of iron to produce 2 moles of iron(III) oxide.

To find the volume of oxygen, we need to convert the given mass of iron(III) oxide to moles and then use the stoichiometry of the reaction to calculate the moles of oxygen required.

The molar mass of iron(III) oxide (Fe2O3) is calculated as follows: (2 x atomic mass of iron) + (3 x atomic mass of oxygen) = (2 x 55.845 g/mol) + (3 x 15.999 g/mol) = 159.69 g/mol

Now, let's calculate the moles of iron(III) oxide: moles of iron(III) oxide = mass / molar mass = 32 g / 159.69 g/mol = 0.2005 mol

According to the stoichiometry of the reaction, 2 moles of iron(III) oxide react with 3 moles of oxygen. Therefore, the moles of oxygen required can be calculated as follows: moles of oxygen = (3/2) x moles of iron(III) oxide = (3/2) x 0.2005 mol = 0.3008 mol

To find the volume of oxygen, we need to use the ideal gas law equation: PV = nRT

Assuming the reaction takes place at standard temperature and pressure (STP), where the temperature is 273.15 K and the pressure is 1 atm, we can rearrange the ideal gas law equation to solve for volume (V): V = (nRT) / P

Substituting the values into the equation: V = (0.3008 mol x 0.0821 L·atm/(mol·K) x 273.15 K) / 1 atm ≈ 6.91 L

Therefore, the volume of oxygen required for the reaction with iron to produce 32g of iron(III) oxide is approximately 6.91 liters.

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