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SrBr2 + (NH4)2CO3 → SrCO3 + 2 NH4Br

If 2.1993 moles of (NH4)2CO3 are reacted, how many grams of SrCO3 will be produced?

324.6g

Explanation:

From this given problem, it is desirable to solve from the known specie to the unknown.

The balanced chemical combination is given as:

          SrBr₂   +     (NH₄)₂CO₃    →     SrCO₃   +    2NH₄Br

Given;

           2.1993 mole of (NH₄)₂CO₃  ;

Unknown:

           Mass of  SrCO₃ produced;

From the balanced equation, we see that;

                1 mole of  (NH₄)₂CO₃ reacted to produce 1 mole of   SrCO₃

                2.1993 mole of (NH₄)₂CO₃ will produce 2.1993 mole of SrCO₃

Mass of SrCO₃ produced = number of moles x molar mass of SrCO₃

Molar mass of SrCO₃ = 87.6 + 12 + 3(16) = 147.6g/mol

Mass of SrCO₃ = 2.1993 x 147.6 = 324.6g