0.1" mole "BaCl_(2)" treated with "0.5" moles "Na_(3)PO_(4)" men no of "Ba_(3)(PO_(4))_(2)" formed? "

Statement-1 : 2 moles of BaCl_(2) react with 3 moles of Na_(3)PO_(4) to form 0.66 mole of Ba_(3)(PO_(4))_(2) Statement-2 : BaCl_(2) is the limiting reactant and some Na_(3)PO_(4) remains unreacted.

If 0.5 mole of BaCl_(2) is mixed with 0.20 mole of Na_(3)PO_(4) , the maximum number of Ba_(3)(PO_(4))_(2) that can be formed is :

If 0.5 mole of BaCl_(2) is mixed with 0.20 mole of Na_(3)PO_(4) , the maximum number of Ba_(3)(PO_(4))_(2) that can be formed is :

If 0.5 mole of BaCl_(2) are mixed with 0.2 mole of Na_(3)PO_(4) ,the maximum number of moles. Of Ba_(3)(PO_(4)_(2)) that can be formed, is

When 0.1 mole of BaCl_(2) is treated with 0.5 mole of Na_(3)PO_(4) , the maximum number of moles of Barium phosphate formed is

If 0.5 mole of BaCl_(2) mixed with 0.20 mole of Na_(3) PO_(4) the maximum number of moles of Ba_(3)(PO)_(2) then can be formed is