The neutralization sulfuric acid with sodium hydroxide yields
HSO4(aq) + Na2OH(aa)
H2SO4(aq) + NaOH(aq)
H2Na(1) + O2SO4(aq)
H2O(l) + Na2SO4(aq)

H2O(l) + Na2SO4(aq)

Explanation:

When sulphuric acid (H2SO4) is neutralized by sodium hydroxide (NaOH), the product formed are sodium sulphate (Na2SO4) and water (H2O). This is illustrated by the equation below:

NaOH + H2SO4 —> Na2SO4 + H20

Now we can balance the equation above as follow:

There are 2 atoms of Na on the right side and 1atom on the left side. It can be balance by putting 2 in front of NaOH as shown below:

2NaOH + H2SO4 —> Na2SO4 + H20

Now we have a total of 4 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 2 in front of H2O as shown below:

2NaOH + H2SO4 —> Na2SO4 + 2H20

Now the equation is balanced.

The product are Na2SO4 and H20

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2