Question
Question: What is the limiting reactant when \( 6.00g \) of \( BaC{l_2} \) reacts with \( 5.00g \) of \( N{a_3...
What is the limiting reactant when 6.00g of BaCl2 reacts with 5.00g of Na3PO4 to form the precipitate Ba3(PO4)2 ?
Solution
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 BaCl2 and Na3PO4 react with each other, and undergoes a double displacement reaction in which
The chemical reaction involved between BaCl2 and Na3PO4 is:
3BaCl2+2Na3PO4→Ba3(PO4)2+6NaCl
Given mass of barium chloride is 6.00g , and molar mass is 208.23gmol−1
Given mass of sodium phosphate is 5.00g , and molar mass is 163.94gmol−1
Moles of BaCl2 will be 208.236=0.0288moles
These are the moles of one molecule, as there were three molecules, nBaCl2=3×0.0288=0.0864mol
Moles of Na3PO4 will be 163.945=0.0305moles
These are the moles of one molecule, as there were two molecules in the reaction to form one molecule of Ba3(PO4)2
nNa3PO4=2×0.0305=0.061mol
As, the moles of Na3PO4 is small, when compared between the moles of BaCl2 and Na3PO4 , it is the limiting reactant.
Note:
Barium chloride is an inorganic compound with the molecular formula of BaCl2 , sodium phosphate is also an inorganic compound with a molecular formula of Na3PO4 . The phosphate and chloride groups were interchanged leading to the formation of Ba3(PO4)2 . The chemical reaction is an example of a double displacement reaction.