Question

Determine the theorical yield (in grams) of strontium sulfate. What is your limiting reactant and excess reactant?
given 2.27×10^-3 mol Na2SO4 and 2.5×10^-3 mol SrCl2

Answer 1

The reaction of Na2SO4 and SrCl2 are following:

Na2SO4 + SrCl2 -- > SrSO4 + 2NaCl

Given that moles of SrCl2: 2.5×10^-3 mol SrCl2

Na2SO4: 2.27×10^-3 mol Na2SO4

SrCl2 due to following reasons:

1. It completely reacted in the reaction.
2. It determines the amount of the product in mol.

2.5×10^-3 mol SrCl2 * 1 mole SrSO4/ 1 Mole SrCl2

2.5×10^-3 mol SrSO4

the theatrical yield (in grams) of strontium sulfate= Number of moles* molar mass

= 2.5×10^-3 mol SrSO4*183.68 g/mol

= 0.4592 g SrSO4