Question

Помогите пожалуйста с химией! Расставить коэффициенты методом электронного баланса,проходящего посхемой; определить окислитель и восстановитель:NH 3+OF 2 = +N 2 +HF+ H2O​

Answer 1

Ответ:

3 OF2 + 4 NH3 → 6 HF + 2 N2 + 3 H2O

Это окислительно-восстановительная (редокс

Объяснение:

извиняюсь если неравильно

Answer 2

Balancing the Chemical Equation

To balance the chemical equation NH3 + OF2 = N2 + HF + H2O using the electron balance method, we need to ensure that the number of atoms of each element is the same on both sides of the equation.

1. Start by counting the number of atoms of each element on both sides of the equation: - On the left side: 1 N, 3 H, 1 O, 1 F - On the right side: 2 N, 1 H, 1 O, 1 F

2. Balance the elements one by one, starting with the elements that appear in the fewest compounds: - Balance the nitrogen (N) atoms by placing a coefficient of 2 in front of NH3 on the left side: 2 NH3 + OF2 = N2 + HF + H2O - Now we have 4 hydrogen (H) atoms on the left side and 2 hydrogen atoms on the right side. Balance the hydrogen atoms by placing a coefficient of 2 in front of HF on the right side: 2 NH3 + OF2 = N2 + 2 HF + H2O - Now we have 2 oxygen (O) atoms on the left side and 1 oxygen atom on the right side. Balance the oxygen atoms by placing a coefficient of 2 in front of H2O on the right side: 2 NH3 + OF2 = N2 + 2 HF + 2 H2O - Finally, we have 2 fluorine (F) atoms on the right side and 1 fluorine atom on the left side. Balance the fluorine atoms by placing a coefficient of 2 in front of OF2 on the left side: 2 NH3 + 2 OF2 = N2 + 2 HF + 2 H2O

Identifying the Oxidizing and Reducing Agents

To determine the oxidizing and reducing agents in the reaction NH3 + OF2 = N2 + 2 HF + 2 H2O, we need to examine the changes in oxidation states of the elements involved.

1. Start by assigning oxidation states to each element in the reactants and products: - In NH3, nitrogen (N) has an oxidation state of -3, and hydrogen (H) has an oxidation state of +1. - In OF2, oxygen (O) has an oxidation state of -2, and fluorine (F) has an oxidation state of -1. - In N2, nitrogen (N) has an oxidation state of 0. - In HF, hydrogen (H) has an oxidation state of +1, and fluorine (F) has an oxidation state of -1. - In H2O, oxygen (O) has an oxidation state of -2, and hydrogen (H) has an oxidation state of +1.

2. Analyze the changes in oxidation states: - In the reaction, the oxidation state of nitrogen (N) changes from -3 in NH3 to 0 in N2. It is gaining electrons and undergoing reduction, so it is the reducing agent. - The oxidation state of oxygen (O) changes from -2 in OF2 to -2 in H2O. It remains the same, so it is not involved in oxidation or reduction. - The oxidation state of fluorine (F) changes from -1 in OF2 to -1 in HF. It remains the same, so it is not involved in oxidation or reduction. - The oxidation state of hydrogen (H) changes from +1 in NH3 to +1 in HF and +1 in H2O. It remains the same, so it is not involved in oxidation or reduction.

Therefore, in the reaction NH3 + OF2 = N2 + 2 HF + 2 H2O, the oxidizing agent is OF2, and the reducing agent is NH3.

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