How many grammes of \(\mathrm{HCl}\) will be required to react with \(10 \mathrm{~g}\) of \(\mathrm{NaOH}\) \((\mathrm{H}=1 . \mathrm{Cl}=1.35 .5 \mathrm{Na}=23, \mathrm{O}=16) ?\)
A. \(9.125 \mathrm{~g}\) B. \(4.6 \mathrm{~g}\) C. \(10.1 \mathrm{~g}\) D. \(18.4 \mathrm{~g}\)
Correct Answer: A
Explanation
\(\mathrm{HCl}+\mathrm{NaOH} \rightarrow \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}\) Mole of \(\mathrm{NaOH}=\frac{10 \mathrm{~g}}{40 \mathrm{gmol}^{-1}}=0.25 \mathrm{~mol}\) Since 1 mole of \(\mathrm{HCl}\) is required for 1 mole of \(\mathrm{NaOH}\) (as shown by the equation above), \(0.25\) mol of \(\mathrm{HCl}\) will be required for \(0.25 \mathrm{~mol}\) of \(\mathrm{NaOH}\). Mass of \(\mathrm{HCl}=(0.25 \times 36.5) \mathrm{g}=9.125 \mathrm{~g}\)
