if .5 mole of baCl2 is mixed with .2 mole of Na3Po4.Then maximum number of mole of Ba3(PO4)2 is
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.
To find the maximum number of moles of Ba3(PO4)2 formed, we need to determine the limiting reactant in the reaction between BaCl2 and Na3PO4. The limiting reactant is the one that gets completely consumed and restricts the amount of product formed.
Step 1: Write the balanced chemical equation for the reaction:
BaCl2(aq) + Na3PO4(aq) → Ba3(PO4)2(s) + 6NaCl(aq)
Step 2: Calculate the number of moles of BaCl2 and Na3PO4 involved in the reaction:
Given: 0.5 mole of BaCl2, 0.2 mole of Na3PO4
Step 3: Calculate the stoichiometric ratio between BaCl2 and Ba3(PO4)2:
From the balanced equation, we can see that 1 mole of BaCl2 reacts with 1 mole of Ba3(PO4)2.
Step 4: Determine the limiting reactant:
To find the limiting reactant, compare the moles of each reactant to their stoichiometric ratio. In this case, we compare the moles of BaCl2 to Ba3(PO4)2.
For BaCl2:
0.5 mole of BaCl2 × (1 mole Ba3(PO4)2 / 1 mole BaCl2) = 0.5 mole Ba3(PO4)2
For Na3PO4:
0.2 mole of Na3PO4 × (1 mole Ba3(PO4)2 / 1 mole Na3PO4) = 0.2 mole Ba3(PO4)2
Since 0.2 mole of Ba3(PO4)2 is less than 0.5 mole of Ba3(PO4)2, Na3PO4 is the limiting reactant.
Step 5: Calculate the maximum number of moles of Ba3(PO4)2 formed:
The maximum number of moles of Ba3(PO4)2 formed is equal to the moles of the limiting reactant, which in this case is 0.2 mole.
Therefore, the maximum number of moles of Ba3(PO4)2 is 0.2 mole.