Question
if .5 mole of baCl2 is mixed with .2 mole of Na3Po4.Then maximum number of mole of Ba3(PO4)2 is
Answers
DrBob222
3Ba^2+ + 2PO4^3- ==> Ba3(PO4)2(s)
Convert mols each to mols product.
0.5 mol BaCl2 x (1 mol product/3 Ba^2+) = approx0.167 mols Ba3(PO4)2 produced.
0.2 mol Na3PO4 x (1 mol product/2 mol PO4^3-) = 0.1 mol Ba3(PO3)2 produced.
In limiting reagent problems the smaller number ALWAYS wins. Therefore, the max number of Ba3(PO4)2 that can be produced is 0.1 mol.
Convert mols each to mols product.
0.5 mol BaCl2 x (1 mol product/3 Ba^2+) = approx0.167 mols Ba3(PO4)2 produced.
0.2 mol Na3PO4 x (1 mol product/2 mol PO4^3-) = 0.1 mol Ba3(PO3)2 produced.
In limiting reagent problems the smaller number ALWAYS wins. Therefore, the max number of Ba3(PO4)2 that can be produced is 0.1 mol.