Respuesta :
Firstly we will convert the mass of the reactants to moles or reactants:
[tex]CaCl_{2(aq)}+Na_2CO_{3(aq)}\rightarrow CaCO_{3(s)}+2NaCl_{(aq)}[/tex][tex]\begin{gathered} _nCaCl_{2(aq)}=\frac{mass}{molar\text{ }mass} \\ _nCaCl_{2(aq)}=\frac{2.0g}{110.98gmol^{-1}} \\ _nCaCl_{2(aq)}=0.0180mol \\ \\ _nNa_2CO_{3(aq)}=\frac{mass}{molar\text{ }mass} \\ _nNa_2CO_{3(aq)}=\frac{2.0g}{105.99gmol^{-1}} \\ _nNa_2CO_{3(aq)}=0.0189mol \end{gathered}[/tex]The easiest way to determine which is the limiting reagent is to determine the moles per co-efficient ratio.
[tex]\begin{gathered} CaCl_{2(aq)}=\frac{0.0180}{1} \\ CaCl_{2(aq)}=0.180 \\ \\ Na_2CO_{3(aq)}=\frac{0.0189}{1} \\ Na_2CO_{3(aq)}=0.0189 \end{gathered}[/tex]Answer: The one with the lowest value is the limiting reactant. CaCl2 has the lowest value and is the limiting reactant.
