18cm^3 of 1.0M H2SO4 just reacted with 24cm^3 of 1.5M NaOH to form sodium sulfate & water. Calculate the amounts in moles of sulfuric acid & sodium hydroxide reacting & write an equation for the reaction.

mols H2SO4 = M x L = 0.018

mols NaOH = M x L = 0.036

2NaOH + H2SO4 ==> Na2SO4 + 2H2O
0.036....0.018....0.018....0.036

To calculate the amounts in moles of sulfuric acid (H2SO4) and sodium hydroxide (NaOH) reacting, we can use the equation:

Molarity (M) = moles (mol) / volume (L)

First, let's calculate the moles of H2SO4:
Moles of H2SO4 = Molarity of H2SO4 × Volume of H2SO4
= 1.0 mol/L × (18 cm³ / 1000 cm³/L)
= 0.018 mol

Next, let's calculate the moles of NaOH:
Moles of NaOH = Molarity of NaOH × Volume of NaOH
= 1.5 mol/L × (24 cm³ / 1000 cm³/L)
= 0.036 mol

Therefore, the moles of sulfuric acid (H2SO4) reacting are 0.018 mol, and the moles of sodium hydroxide (NaOH) reacting are 0.036 mol.

Now, let's write the balanced equation for the reaction:

H2SO4 + 2NaOH → Na2SO4 + 2H2O

This equation shows that one molecule of sulfuric acid (H2SO4) reacts with two molecules of sodium hydroxide (NaOH), producing one molecule of sodium sulfate (Na2SO4) and two molecules of water (H2O).