balance equation for sodium sulfate + hydrochloric acid = sodium chloride + water +Sulfur dioxide ?
I prefer to do it this way.
Na2SO4 + 2HCl ==> 2NaCl + H2SO4
Then H2SO4 decomposes by
H2SO4 ==> H2O + SO3
Then SO3 decomposes by
2SO3 --> 2SO2 + O2
You can put all of this together as
2Na2SO4 + 4HCl ==> 4NaCl + 2H2O + 2SO2 + O2
To balance the equation for the reaction between sodium sulfate (Na2SO4) and hydrochloric acid (HCl) to form sodium chloride (NaCl), water (H2O), and sulfur dioxide (SO2), follow these steps:
Step 1: Write down the unbalanced equation:
Na2SO4 + HCl → NaCl + H2O + SO2
Step 2: Count the number of atoms of each element on both sides of the equation. This will help you determine which coefficients are needed to balance the equation.
Left side:
Na: 2
S: 1
O: 4
H: 1
Cl: 1
Right side:
Na: 1
S: 1
O: 2
H: 2
Cl: 1
Step 3: Start balancing the equation by adjusting the coefficients.
Start with the sodium (Na) atoms. Since there are 2 Na atoms on the left side and only 1 Na atom on the right side, place a coefficient of 2 in front of NaCl:
Na2SO4 + HCl → 2NaCl + H2O + SO2
Next, balance the sulfur (S) atoms. Since there is 1 S atom on both sides, no further adjustment is needed:
Na2SO4 + HCl → 2NaCl + H2O + SO2
Now, balance the chlorine (Cl) atoms. There is 1 Cl atom on both sides, so no further adjustment is needed:
Na2SO4 + 2HCl → 2NaCl + H2O + SO2
Next, balance the oxygen (O) atoms. There are 4 O atoms on the left side, and only 3 O atoms on the right side. To balance the O atoms, add a coefficient of 2 in front of H2O:
Na2SO4 + 2HCl → 2NaCl + 2H2O + SO2
Finally, balance the hydrogen (H) atoms. There are 2 H atoms on the right side, so add a coefficient of 4 in front of HCl:
Na2SO4 + 4HCl → 2NaCl + 2H2O + SO2
Step 4: Verify that the equation is balanced by counting the atoms again:
Left side:
Na: 2
S: 1
O: 4
H: 4
Cl: 4
Right side:
Na: 2
S: 1
O: 4
H: 4
Cl: 4
Now the equation is balanced, and the balanced equation for the reaction is:
Na2SO4 + 4HCl → 2NaCl + 2H2O + SO2