Question

Расставьте коэффициенты, найдите окислитель и востовитель. K2Cr2O7+Na2SO3+H2SO4=K2SO4+Cr2(SO4)3+Na2SO4+H2O. Cu2O+HNO3=Cu(NO3)2+NO2+H2O помогите

Answer 1

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Answer 2

Balancing the Chemical Equations

To balance the given chemical equations, we need to ensure that the number of atoms of each element is the same on both sides of the equation. Let's start with the first equation:

Equation 1: K2Cr2O7 + Na2SO3 + H2SO4 = K2SO4 + Cr2(SO4)3 + Na2SO4 + H2O

To balance this equation, we can follow these steps:

Step 1: Balance the atoms other than hydrogen and oxygen. - We have 2 potassium (K) atoms on the left side, so we need 2 potassium atoms on the right side. We can achieve this by placing a coefficient of 2 in front of K2SO4. - We have 2 sodium (Na) atoms on the left side, so we need 2 sodium atoms on the right side. We can achieve this by placing a coefficient of 2 in front of Na2SO4. - We have 2 chromium (Cr) atoms on the left side, so we need 2 chromium atoms on the right side. We can achieve this by placing a coefficient of 2 in front of Cr2(SO4)3.

The equation now becomes: K2Cr2O7 + Na2SO3 + H2SO4 = 2K2SO4 + 2Cr2(SO4)3 + 2Na2SO4 + H2O

Step 2: Balance the oxygen atoms. - We have 7 oxygen (O) atoms on the left side, so we need 7 oxygen atoms on the right side. We can achieve this by placing a coefficient of 7 in front of H2O.

The equation now becomes: K2Cr2O7 + Na2SO3 + H2SO4 = 2K2SO4 + 2Cr2(SO4)3 + 2Na2SO4 + 7H2O

Step 3: Balance the hydrogen atoms. - We have 2 hydrogen (H) atoms on the left side, so we need 14 hydrogen atoms on the right side. We can achieve this by placing a coefficient of 14 in front of H2SO4.

The final balanced equation is: K2Cr2O7 + Na2SO3 + 14H2SO4 = 2K2SO4 + 2Cr2(SO4)3 + 2Na2SO4 + 7H2O

Now let's move on to the second equation:

Equation 2: Cu2O + HNO3 = Cu(NO3)2 + NO2 + H2O

To balance this equation, we can follow these steps:

Step 1: Balance the atoms other than hydrogen and oxygen. - We have 2 copper (Cu) atoms on the left side, so we need 2 copper atoms on the right side. We can achieve this by placing a coefficient of 2 in front of Cu(NO3)2.

The equation now becomes: Cu2O + HNO3 = 2Cu(NO3)2 + NO2 + H2O

Step 2: Balance the oxygen atoms. - We have 1 oxygen (O) atom on the left side, so we need 6 oxygen atoms on the right side. We can achieve this by placing a coefficient of 6 in front of HNO3.

The equation now becomes: Cu2O + 6HNO3 = 2Cu(NO3)2 + NO2 + H2O

Step 3: Balance the hydrogen atoms. - We have 6 hydrogen (H) atoms on the left side, so we need 12 hydrogen atoms on the right side. We can achieve this by placing a coefficient of 12 in front of H2O.

The final balanced equation is: Cu2O + 6HNO3 = 2Cu(NO3)2 + NO2 + 12H2O

Identifying the Oxidizing and Reducing Agents

In the first equation, the oxidizing agent is K2Cr2O7 (potassium dichromate), which is being reduced. The reducing agent is Na2SO3 (sodium sulfite), which is being oxidized.

In the second equation, the oxidizing agent is HNO3 (nitric acid), which is being reduced. The reducing agent is Cu2O (copper(I) oxide), which is being oxidized.

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