Question

Закончить уравнение реакции а) и расставить коэффициенты реакциях а) и б). a) Mg + HNO3(конц) = б) NaI + KMnO4 + H2SO4 = K2SO4 + MnSO4 + Na2SO4 + I2 + H2O.

Answer 1

.....,........,.........

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 balance the equations step by step.

a) Mg + HNO3 (конц) = ?

First, let's balance the equation for the reaction between magnesium (Mg) and nitric acid (HNO3).

The unbalanced equation is:

Mg + HNO3 → ?

To balance the equation, we start by counting the number of atoms of each element on both sides of the equation.

On the left side: - Magnesium (Mg): 1 atom - Nitrogen (N): 1 atom - Oxygen (O): 3 atoms (1 from HNO3)

On the right side: - Magnesium (Mg): 1 atom - Nitrogen (N): 1 atom - Oxygen (O): 3 atoms (1 from HNO3)

Since the number of atoms of each element is already the same on both sides, the equation is already balanced.

The balanced equation is:

Mg + HNO3 → Mg(NO3)2 + H2

b) NaI + KMnO4 + H2SO4 = ?

Next, let's balance the equation for the reaction between sodium iodide (NaI), potassium permanganate (KMnO4), and sulfuric acid (H2SO4).

The unbalanced equation is:

NaI + KMnO4 + H2SO4 → ?

To balance the equation, we start by counting the number of atoms of each element on both sides of the equation.

On the left side: - Sodium (Na): 1 atom - Iodine (I): 1 atom - Potassium (K): 1 atom - Manganese (Mn): 1 atom - Oxygen (O): 4 atoms (4 from KMnO4) - Hydrogen (H): 2 atoms (2 from H2SO4) - Sulfur (S): 1 atom (1 from H2SO4)

On the right side: - Sodium (Na): 2 atoms (2 from NaI) - Iodine (I): 2 atoms (2 from NaI) - Potassium (K): 2 atoms (2 from KMnO4) - Manganese (Mn): 1 atom - Oxygen (O): 12 atoms (4 from KMnO4, 8 from H2SO4) - Hydrogen (H): 4 atoms (2 from H2SO4, 2 from H2O) - Sulfur (S): 4 atoms (1 from H2SO4, 3 from K2SO4)

To balance the equation, we can start by balancing the atoms that appear in only one compound on each side. In this case, we can balance the iodine (I) atoms by placing a coefficient of 2 in front of NaI:

2 NaI + KMnO4 + H2SO4 → ?

Now, let's count the atoms again:

On the left side: - Sodium (Na): 2 atoms (2 from NaI) - Iodine (I): 2 atoms (2 from NaI) - Potassium (K): 1 atom - Manganese (Mn): 1 atom - Oxygen (O): 4 atoms (4 from KMnO4) - Hydrogen (H): 2 atoms (2 from H2SO4) - Sulfur (S): 1 atom (1 from H2SO4)

On the right side: - Sodium (Na): 2 atoms (2 from NaI) - Iodine (I): 2 atoms (2 from NaI) - Potassium (K): 2 atoms (2 from KMnO4) - Manganese (Mn): 1 atom - Oxygen (O): 12 atoms (4 from KMnO4, 8 from H2SO4) - Hydrogen (H): 4 atoms (2 from H2SO4, 2 from H2O) - Sulfur (S): 4 atoms (1 from H2SO4, 3 from K2SO4)

To balance the remaining atoms, we can place coefficients in front of KMnO4, H2SO4, and K2SO4:

2 NaI + 5 KMnO4 + 8 H2SO4 → K2SO4 + 5 MnSO4 + Na2SO4 + 4 I2 + 8 H2O

The balanced equation is:

2 NaI + 5 KMnO4 + 8 H2SO4 → K2SO4 + 5 MnSO4 + Na2SO4 + 4 I2 + 8 H2O

Please note that the coefficients in front of the compounds represent the number of molecules or formula units, not the number of atoms.