Answer:
3.94 × 10³ g
Explanation:
Step 1: Write the neutralization reaction
HClO₃ + NaOH = NaClO₃ + H₂O
Step 2: Calculate the moles of chloric acid that react
25.5 L of 3.86 M chloric acid is to be neutralized. The reacting moles are:
[tex]25.5L \times \frac{3.86mol}{L} =98.4mol[/tex]
Step 3: Calculate the required moles of sodium hydroxide
The molar ratio of HClO₃ to NaOH is 1:1. Then, the reacting moles of NaOH are 98.4 moles.
Step 4: Calculate the mass of sodium hydroxide corresponding to 98.4 moles
The molar mass of NaOH is 40.00 g/mol.
[tex]98.4mol \times \frac{40.00g}{mol} =3.94 \times 10^{3} g[/tex]
