Question

How many grams of Ba₃(PO₄)₂ can be made when 100 mL of 0.300 M BaCl₂ are allowed to react with 50.0 mL of 0.800 M Na₃PO₄?

Answer 1

first we should write the balanced equation

3BaCl2 + 2Na3PO4 ==> 6NaCl + Ba3(PO4)2

first calculate the moles of BaCl2 and Na3PO4 using

no of moles = molarity x volume in liters (keep in mind you have to change volume in to liters)

moles of BaCl2 = 0.3 x 0.1 L

= 0.03 moles of BaCl2

no of moles of Na3PO4 = 0.8 x 0.05

= 0.04 moles

from the balanced equation it is clear that to produce one mole of Ba3(PO4)2 we need 3 mole of BaCl2 and two mole Na3PO4

but here we have more of more of Na3PO4 so limiting agent is BaCl2

so from two 3 moles of BaCl2 we will get one mole of Ba3(PO4)2

0.03 moles of BaCl3 how many moles Ba3(PO4)3 will come

= 0.03 / 3

= 0.01 moles Ba3(PO)4 will come

no we know the moles we can calculte the grams using

weight = no of moles x molar mass

= 0.01 x 603.6226

= 6.0362 grams