Question

When reacting 2.0g of CaCl2 with 2.0 g of Na2CO3, what was the limiting reagent based on the following equation?CaCl2 (aq) + Na2CO3 (aq) --> CaCO3 (s) + 2 NaCl (aq)

Answer 1

Firstly we will convert the mass of the reactants to moles or reactants:

`CaCl_(2(aq))+Na_2CO_(3(aq))\rightarrow CaCO_(3(s))+2NaCl_((aq))`
`\begin{gathered} _nCaCl_(2(aq))=\frac{mass}{molar\text{ }mass} \\ _nCaCl_(2(aq))=(2.0g)/(110.98gmol^(-1)) \\ _nCaCl_(2(aq))=0.0180mol \\ \\ _nNa_2CO_(3(aq))=\frac{mass}{molar\text{ }mass} \\ _nNa_2CO_(3(aq))=(2.0g)/(105.99gmol^(-1)) \\ _nNa_2CO_(3(aq))=0.0189mol \end{gathered}`

The easiest way to determine which is the limiting reagent is to determine the moles per co-efficient ratio.

`\begin{gathered} CaCl_(2(aq))=(0.0180)/(1) \\ CaCl_(2(aq))=0.180 \\ \\ Na_2CO_(3(aq))=(0.0189)/(1) \\ Na_2CO_(3(aq))=0.0189 \end{gathered}`

Answer: The one with the lowest value is the limiting reactant. CaCl2 has the lowest value and is the limiting reactant.