Answer
$4.42g\ HCl$
Work Step by Step
$4.85g\ NaOH\times\frac{1mol\ NaOH}{39.997g\ NaOH}\times\frac{1mol\ HCl}{1mol\ NaOH}\times\frac{36.46g\ HCl}{1mo l\ HCl}= 4.42g\ HCl$
We divide the mass in grams of the base by its molar mass to get the number of moles of base. We then multiply this by the ratio obtained from the balanced chemical equation with the base on the bottom and the acid on the top. We now have the number of moles of acid necessary to neutralize the base. We multiply this by the molar mass of the acid to get our result.
