Question

Используя метод электронного баланса, составьте уравнение реакции по схеме. Определите окислитель и восстановитель. Расставьте коэффициенты HI + H2SO4 → H2S + I2 + H2O P + H2SO4 → P2O5 + SO2 + H2O СРОЧНО!!!

Answer 1

Ответ:

ну вот как-то так получилось

надеюсь 2 не будет

Answer 2

Equation 1: HI + H2SO4 → H2S + I2 + H2O

To balance the equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation.

Let's start by balancing the hydrogen (H) atoms: - On the left side, we have 2 hydrogen atoms from HI and 4 hydrogen atoms from H2SO4, giving us a total of 6 hydrogen atoms. - On the right side, we have 2 hydrogen atoms from H2S and 2 hydrogen atoms from H2O, giving us a total of 4 hydrogen atoms.

To balance the hydrogen atoms, we can add a coefficient of 2 in front of H2O on the right side: HI + H2SO4 → H2S + I2 + 2H2O

Next, let's balance the sulfur (S) atoms: - On the left side, we have 1 sulfur atom from H2SO4. - On the right side, we have 1 sulfur atom from H2S.

The sulfur atoms are already balanced.

Now, let's balance the iodine (I) atoms: - On the left side, we have 1 iodine atom from HI. - On the right side, we have 1 iodine atom from I2.

The iodine atoms are already balanced.

The balanced equation is: 2HI + H2SO4 → H2S + I2 + 2H2O

In this reaction, HI is the reducing agent (or the reductant) because it donates electrons and gets oxidized, while H2SO4 is the oxidizing agent (or the oxidant) because it accepts electrons and gets reduced.

Equation 2: P + H2SO4 → P2O5 + SO2 + H2O

Let's balance the equation by ensuring that the number of atoms of each element is the same on both sides.

Starting with the hydrogen (H) atoms: - On the left side, we have 2 hydrogen atoms from H2SO4. - On the right side, we have 2 hydrogen atoms from H2O.

The hydrogen atoms are already balanced.

Next, let's balance the sulfur (S) atoms: - On the left side, we have 1 sulfur atom from H2SO4. - On the right side, we have 1 sulfur atom from SO2.

The sulfur atoms are already balanced.

Now, let's balance the phosphorus (P) atoms: - On the left side, we have 1 phosphorus atom from P. - On the right side, we have 2 phosphorus atoms from P2O5.

To balance the phosphorus atoms, we can add a coefficient of 2 in front of P on the left side: 2P + H2SO4 → P2O5 + SO2 + H2O

In this reaction, P is the reducing agent (or the reductant) because it donates electrons and gets oxidized, while H2SO4 is the oxidizing agent (or the oxidant) because it accepts electrons and gets reduced.

Please note that these equations and classifications of oxidizing and reducing agents are based on the information provided and may vary depending on the specific conditions of the reaction.