Question

Na2SO4(mL)	BaCl2(mL)	BaSO4(g) Actual Grams	BaSO4(g) Actual Grams
8	2	0.58	0.54
8	4	1.59	2.03
8	6	2.45	2.39
8	8	1.97	2.13
8	10	1.61	2.08
8	12	1.84	2.59
				Can someone please help me in calculating the theoretical grams from this information. thanks.

Answer 1

Step 1. Calculate the no. of moles of each reactant. ( I am showing the 1st example)

Na2SO4 is taken 8ml

density = mass/ volume

density of Na2SO4 =  2.68 g/mL= mass / volume

mass = 2.68 g/mL x 8 mL = 21.44 g

no. of moles of Na2SO4 = mass taken/molar mass = 21.44/142.04 g/mol = 0.15 moles

For BaCl2,

density = 3.856 g/mL = mass/2mL

mass = 3.856g/mL x 2mL = 7.712 g

no. of moles = 7.712/208.23g/mol = 0.037 moles

Step 2: know the limiting reagent, means the reactant which is less and due to which the reaction do not proceed furthur.

Na2SO4 is 0.15 moles

BaCl2 is 0.037 moles

So, limiting reagent is BaCl2 with 0.037 moles

Step 3: Estimate the no.of moles of product formed due to limiting reagent

By the reaction, Na2SO4 + BaCl2 ---> BaSO4 + 2 NaCl

means 1 mole BaCl2 produces 1 mole BaSO4

In our example, 0.037 moles BaCl2 produces 0.037 moles BaSO4

Step 4: know the theoretical yield

Now, 0.037 moles of product formed

molar mass of BaSO4 =233.43g/mol

so theoretical ield = mass produced = no.of moles x molar mass = 0.037mol x 233.43g/mol = 8.64