Question

Na2co3-->nacl-->na-->nai-->nano3-->na2cro4

Answer 1

Na2CO3+2HCl=2NaCl+H2O+CO2
2NaC=Na+Cl2
2Na+I2=2NaI
2NaI(тверд.) + 4HNO3(конц.)=I2+2NO2+2H2O+2NaNO3
3NaNO3+4NaOH+Cr2O3=2Na2CrO4+3NaNO2+2H2O

Answer 2

Chemical Reaction: Na2CO3 -> NaCl -> Na -> NaI -> NaNO3 -> Na2CrO4

The given chemical reaction involves the transformation of sodium carbonate (Na2CO3) into sodium chromate (Na2CrO4) through a series of intermediate steps. Let's break down the reaction step by step:

1. Sodium Carbonate (Na2CO3) to Sodium Chloride (NaCl): - The first step in the reaction is the decomposition of sodium carbonate (Na2CO3) into sodium chloride (NaCl). - This can be represented by the following equation: Na2CO3 -> NaCl. - The reaction involves the exchange of the carbonate ion (CO3^2-) with the chloride ion (Cl^-).

2. Sodium Chloride (NaCl) to Sodium (Na): - In the next step, sodium chloride (NaCl) is further decomposed into sodium (Na). - This can be represented by the following equation: NaCl -> Na. - The reaction involves the displacement of the chloride ion (Cl^-) with the sodium ion (Na+).

3. Sodium (Na) to Sodium Iodide (NaI): - The next step involves the reaction of sodium (Na) with iodine (I2) to form sodium iodide (NaI). - This can be represented by the following equation: Na + I2 -> NaI. - The reaction involves the combination of sodium (Na) with iodine (I2) to form sodium iodide (NaI).

4. Sodium Iodide (NaI) to Sodium Nitrate (NaNO3): - In the subsequent step, sodium iodide (NaI) reacts with nitric acid (HNO3) to form sodium nitrate (NaNO3). - This can be represented by the following equation: NaI + HNO3 -> NaNO3 + HI. - The reaction involves the displacement of iodine (I) with the nitrate ion (NO3^-).

5. Sodium Nitrate (NaNO3) to Sodium Chromate (Na2CrO4): - Finally, sodium nitrate (NaNO3) reacts with sodium chromate (Na2CrO4) to form sodium chromate (Na2CrO4). - This can be represented by the following equation: NaNO3 + Na2CrO4 -> Na2CrO4 + NaNO3. - The reaction involves the exchange of the nitrate ion (NO3^-) with the chromate ion (CrO4^2-).

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