Question

If you start with ten grams of sodium hydroxide , how many moles of sodium chloride will be produced?
\[NaO{H_{(aq)}} + HC{l_{(aq)}} \to NaC{l_{(aq)}} + {H_2}{O_{(l)}}\]

Answer 1

Hint: For dealing with this question, first of all we need to know that we should start it by converting the mass of sodium hydroxide to its respective moles. Then there will be existing a particular mole ratio which exists between those two reactants. Next we have to keep in mind that this will further determine how many moles of chlorine to be produced which is asked as in the above question. We need a clear idea about each and every step taking place here.

Complete step-by-step answer:
First of all our aim is to convert those masses of sodium hydroxide to its own particular number of moles. Here the mass given is ten gram. Molar mass of sodium hydroxide can be taken approximately as \[40.0g\]. So we can find the number of moles of \[NaOH\] in the following way,
\[10g \times \dfrac{{1moleNaOH}}{{40.0g}} = 0.25mole\] of \[NaOH\]

This also allows us to determine \[1:1mole\] ratio which is existing between the two reactants.
Here , in the above chemical reaction given, it is given that \[1\] mole of sodium hydroxide reacts with one mole of hydrochloric acid to form one mole of sodium chloride in aqueous state.

As we know the problem given is not mentioning the particular quantity of hydrochloric acid which is available. So , it enables us to assume that the hydrochloric acid may be in an excess condition so as to consume all the moles of sodium hydroxide. This determines that the reaction will produce as follows,
\[0.25molesNaOH \times \dfrac{{1moleNaCl}}{{1moleNaOH}}\]
\[ = 0.25moleNaCl\]

By assuming the mass of sodium hydroxide to be \[10g\], we need to round the answer to one significant figure. So it will be \[0.3moles\] .

Hence , number of moles of sodium chloride produce will be \[0.3moles\]

Note: Here we have to know that the given reaction is a neutralization reaction. The calculations should not start without converting the mass of sodium hydroxide to the number of moles as it can only help in doing further steps. We will be a bit confused with the given problem as there is no quantity of hydrochloric acid given , in this condition the only way to move further is to assume the quantity to be in excess. Rounding of the final answer should be done related to the mass given.