Question

How many grams of solid is produced when 50 ml of 0.2 mL of 0.2 M Na2CO3 is mixed with 50 mL of 0.158 M of BaCl2.

Answer 1

Na2CO3 (Aq) + BaCl2(aq) ----> 2 NaCl (aq) + BaCO3(s)
this is the balanced reaction where solid BaCO3 will get precipitated We need

mass of this BaCO3 formed for that ,let us find out moles of reactants present
in the solution.And then find out limiting reagent.

given :

volume of Na2CO3 = 50 ml = 0.05 L

molarity of Na2CO3 = 0.2 M

So # moles of Na2CO3 present = volume * molarity
= 0.05 l *0.2 M = 0.01 moles

volume of BaCl2 = 50 ml = 0.05 L
molarity = 0.158 M
thus # moles of BaCl2 = 0.05 L*0.158 M = 0.0079 moles

Since # moles of BaCL2 is less than # moles of Na2CO3,the limiting reactant
is BaCL2.That means when we calculate number of moles of BaCO3 formed ,we
need to consider moles of BaCL2 only.
from balanced equation it is clear that 1 mole BaCL2 produce 2 mole BaCO3
mole ratio 1:2

0.0079 moles BaCL2 produce 2*0.0079 mole BaCO3
thus moles of BaCO3 formed = 0.0158 mols

now convert the moles of BaCO3 in to mass in grams using molar mass

Molar mass of BaCO3 =1* 137.33 + 1*12.01 + 3*16 =197.34 g/mol

1 mole BaCO3 = 197.34 g

therefore 0.0158 mols BaCO3 = 0.0158 mols*197.34 g/mol =3.1179 g

Thus mass of BaCO3 formed = 3.1179 g
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hope it is helpful :)