Question

Расставить коэффициенты в окислительно-восстановительной реакции методом электронного баланса:HNO3+Mg->Mg(HNO3)2+N2O+H2O

Answer 1

Может не Mg(HNO3)2,а Mg(NO3)2?

 H(+)N(5+)O3(2-)+Mg(0)===>Mg(2+)(N(5+)O3(2-))2+N2(+)O(2-)+H2(+)O(2-)

Все знаки и цифры в скобках пишем над элементами. 

Mg - 2e = Mg(+2)          4 ,восстановитель,оксисление
2N(+5) + 8e = 2N(+1)     1,окислитель,восстановление

Расставляем:

10HNO3+4Mg===> 4Mg(NO3)2+N2O+5H2O

 

Answer 2

Balancing the Redox Reaction: HNO3 + Mg -> Mg(HNO3)2 + N2O + H2O

To balance the given redox reaction, HNO3 + Mg -> Mg(HNO3)2 + N2O + H2O, we will use the method of electron balance. This method involves balancing the number of electrons gained and lost during the reaction.

Let's break down the reaction into half-reactions:

Oxidation Half-Reaction: Mg -> Mg(HNO3)2

Reduction Half-Reaction: HNO3 -> N2O + H2O

Now, let's balance the half-reactions one by one:

Balancing the Oxidation Half-Reaction: Mg -> Mg(HNO3)2

In the oxidation half-reaction, magnesium (Mg) is oxidized from an oxidation state of 0 to +2 in Mg(HNO3)2. This means that Mg loses 2 electrons.

To balance the number of electrons, we need to multiply the oxidation half-reaction by 2:

2Mg -> 2Mg(HNO3)2

Balancing the Reduction Half-Reaction: HNO3 -> N2O + H2O

In the reduction half-reaction, nitrogen in HNO3 undergoes a change in oxidation state from +5 to +2 in N2O. This means that nitrogen gains 3 electrons.

To balance the number of electrons, we need to multiply the reduction half-reaction by 3:

3HNO3 -> N2O + 3H2O

Now, let's combine the balanced half-reactions:

2Mg + 3HNO3 -> 2Mg(HNO3)2 + N2O + 3H2O

Therefore, the balanced equation for the given redox reaction is:

2Mg + 3HNO3 -> 2Mg(HNO3)2 + N2O + 3H2O.

Please note that this is a balanced equation based on the given reaction. If there are any additional conditions or constraints, the coefficients may need to be adjusted accordingly.