Question

42 ml of 1.1 mol/L hydrochloric acid reacted with excess sodium hydroxide. How many moles of sodium chloride were produced?

Answer 1

The balanced reaction equation is:

HCl + NaOH $\rightarrow$ H₂O + NaCl.

As one can see, when 1 mol of hydrochloric acid reacts , 1 mol of sodium chloride is produced:

$n(HCl)=n(NaCl)$ .

The number of the moles of hydrochloric acid reacted is the product of the volume and the concentration of the HCl solution:

$n(HCl) = cV = 1.1(\text{mol/L)}·42·10^{-3}(\text{L}) = 0.0462$ mol.

Therefore, the number of the moles of NaCl formed:

$n(NaCl) = n(HCl) = 0.0462$ mol.

Answer: when 42 ml of 1.1 mol/L hydrochloric acid reacts with excess sodium hydroxide, 0.0462 mol of NaCl is produced.