Question

Оформите задачи по химии 1)C2H2 + 2.5O2= 2CO2+ H2O раставим коэф. найдем моляр. массу С2Н2= 26г/моль найдем моляр. массу диоксила углерода = 88г/моль составим пропорцию: 260г-----Х 26г/моль----88г/моль отсюда Х=880гр 2)C2H2 + 2.5O2= 2CO2+ H2O раставим коэф. найдем моляр. массу С2Н2= 26г/моль найдем моляр. массу диоксила углерода = 88г/моль составим пропорцию: 260г-----Х

Answer 1

Вчера решал ее
2С2Н2 + 5О2 = 4СО2 + 2Н2О
V 2                      4
M 26г/моль       44г/моль
m 52г                176г
 260 ----------------х
 52------------------176
х =880г
коэфиценты я другие поставил без дробей

Answer 2

Balancing the Chemical Equation:

The given chemical equation is: C2H2 + 2.5O2 = 2CO2 + H2O.

To balance this equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation. Here's how we can balance it:

C2H2 + 2.5O2 = 2CO2 + H2O

To balance the oxygen atoms, we can multiply the coefficient of O2 by 2:

C2H2 + 5O2 = 2CO2 + H2O

Now, let's balance the carbon atoms. We have 2 carbon atoms on the left side and 2 carbon atoms on the right side, so they are already balanced.

Finally, let's balance the hydrogen atoms. We have 2 hydrogen atoms on the left side and 2 hydrogen atoms on the right side, so they are also balanced.

The balanced equation is: C2H2 + 5O2 = 2CO2 + H2O.

Calculating the Molar Mass:

To calculate the molar mass of a compound, we need to sum up the atomic masses of all the atoms in the compound.

The molar mass of C2H2 (acetylene) is 26 g/mol [[1]]. The molar mass of CO2 (carbon dioxide) is 44 g/mol [[2]].

Using Proportions to Find the Mass:

To find the mass of a substance using proportions, we can set up a ratio of the molar masses of the substances involved.

Given: Mass of C2H2 = 260 g Molar mass of C2H2 = 26 g/mol Molar mass of CO2 = 44 g/mol

Using the proportion: 260 g ----- X 26 g/mol ---- 44 g/mol

Solving for X, we find: X = (260 g * 44 g/mol) / 26 g/mol = 440 g

Therefore, the mass of CO2 produced when 260 g of C2H2 reacts is 440 g.

Please let me know if there's anything else I can help you with!