Question

What is the limiting reactant when $ 6.00g $ of $ BaC{l_2} $ reacts with $ 5.00g $ of $ N{a_3}P{O_4} $ to form the precipitate $ B{a_3}{\left( {P{O_4}} \right)_2} $ ?

Answer 1

Hint: The number of moles can be calculated from the given mass and molar mass. The obtained moles are for one molecule of compound. By multiplying the obtained moles with the number of molecules in the balanced chemical equation gives the total moles involved in the reaction. The molecule with less moles is the limiting reactant.

Complete Step By Step Answer:
Molecules are the combination of atoms. Two or more molecules reacting with each other leads to the formation of new products. There are different types of chemical reactions like single displacement, and double displacement reactions.
Both the compounds $ BaC{l_2} $ and $ N{a_3}P{O_4} $ react with each other, and undergoes a double displacement reaction in which
The chemical reaction involved between $ BaC{l_2} $ and $ N{a_3}P{O_4} $ is:
 $ 3BaC{l_2} + 2N{a_3}P{O_4} \to B{a_3}{\left( {P{O_4}} \right)_2} + 6NaCl $
Given mass of barium chloride is $ 6.00g $ , and molar mass is $ 208.23gmo{l^{ - 1}} $
Given mass of sodium phosphate is $ 5.00g $ , and molar mass is $ 163.94gmo{l^{ - 1}} $
Moles of $ BaC{l_2} $ will be $ \dfrac{6}{{208.23}} = 0.0288moles $
These are the moles of one molecule, as there were three molecules, $ {n_{BaC{l_2}}} = 3 \times 0.0288 = 0.0864mol $
Moles of $ N{a_3}P{O_4} $ will be $ \dfrac{5}{{163.94}} = 0.0305moles $
These are the moles of one molecule, as there were two molecules in the reaction to form one molecule of $ B{a_3}{\left( {P{O_4}} \right)_2} $
 $ {n_{N{a_3}P{O_4}}} = 2 \times 0.0305 = 0.061mol $
As, the moles of $ N{a_3}P{O_4} $ is small, when compared between the moles of $ BaC{l_2} $ and $ N{a_3}P{O_4} $ , it is the limiting reactant.

Note:
Barium chloride is an inorganic compound with the molecular formula of $ BaC{l_2} $ , sodium phosphate is also an inorganic compound with a molecular formula of $ N{a_3}P{O_4} $ . The phosphate and chloride groups were interchanged leading to the formation of $ B{a_3}{\left( {P{O_4}} \right)_2} $ . The chemical reaction is an example of a double displacement reaction.