Answer:
[tex]NH_4Cl[/tex]
Explanation:
Let's write the reaction occurring upon mixing calcium hydroxide and ammonium chloride:
[tex]Ca(OH)_2 (aq)+2 NH_4Cl(aq)\rightarrow CaCl_2(aq)+2 NH_3(g)+2 H_2O(l)[/tex]
Now let's find moles of each reactant. In order to find moles, let's divide mass of each reactant by its molar mass:
[tex]n_{Ca(OH)_2}=\frac{7 g}{74.093 \frac{g}{mol}} =0.0945 mol\\n_{NH_4Cl}=\frac{7 g}{53.491 \frac{g}{mol}}= 0.131 mol[/tex]
Now, to find the limiting reagent and the reagent in excess, we need to take these amounts and divide by the stoichiometric coefficients present in the equation to find equivalents. Calcium hydroxide has a coefficient of '1', while ammonium chloride has '2', therefore:
[tex]eq._{Ca(OH)_2}=\frac{0.0945 mol}{1} =0.0945 mol[/tex]
[tex]eq._{NH_4Cl}=\frac{0.131 mol}{2} =0.0655 mol[/tex]
When we have our equivalents, we find the lower one to identify our limiting reactant. Ammonium chloride has a lower equivalent than calcium hydroxide, so it's our limiting reagent, while a reagent with a higher equivalent is the one in excess.
