Lets consider the reaction of aqueous solutions of strontium

Let’s consider the reaction of aqueous solutions of strontium bromide and ammonium carbonate. Also consider that you begin with 50 g of strontium bromide and 50 g of ammonium carbonate. If you are able to isolate 15 g of strontium carbonate, lets work through determining the percent yield and how much excess reagent is leftover after the reaction. Let’s say that all of your strontium bromide reacted to form strontium carbonate, what is the theoretical yield (in grams) of strontium carbonate? In grams.

Solution

Molar mass of SrBr2 = 247.4 g/mol

Molar mass of (NH4)2CO3 = 96 g/mol

The balanced equation is

strontium bromide + ammonium carbonate --------------------> strontium carbonate

SrBr2 + (NH4)2CO3 -------------> SrCO3 + 2NH4Br

Moles of SrBr2 = mass/ molar mass = 50g / 247.4 g/mol = 0.202 mol

Moles of (NH4)2CO3 = mass/ molar mass = 50g / 96 g/mol = 0.52 mol

Moles of SrBr2 are less compted to (NH4)2CO3.

Therefore, SrBr2 is the Limiting reagent and  (NH4)2CO3 is the Excess reagent.

Theoretical yield is calculated based on the Limiting reagent.

SrBr2 + (NH4)2CO3 -------------> SrCO3 + 2NH4Br

1 mol 1 mol

247.4 g 147.6 g

50 g ?

Then,

? = ( 50 g/ 247.4 g) x 147.6 g of SrCO3

= 29.83 g of SrCO3

Therefore,

Theoretical yield of SrCO3 = 29.83 g

--------------------------------------------------------------------------

Excess moles of (NH4)2CO3 = 0.52 mol - 0.202 mol

= 0.318 mol

Excess mass of (NH4)2CO3 = excess moles x molar mass

= 0.318 mol x 96 g/mol

= 30.53 g

Therefore,

30.53 g of excess reagent is leftover after the reaction