When 214.5 g of calcium carbonate react with 321.9 g of aluminum fluoride, how many grams of each product can be produced?

grams CaF₂ = grams Al₂(CO₃)₃ = 167.7 grams (4 sig. figs.)

Explanation:

                     3CaCO₃          +           2AlF₃            =>     3CaF₂   +   Al₂(CO₃)₃

Given:    214.5g/(100g/mol)      321.9g/(84g/mol) =>2.15 moles & 0.71 mole

                   = 2.15 mole               = 3.83 mole    =>       ?g CaF₂    ?g Al₂(CO₃)₃

ID of Limiting Reactant => divide mole values by respective coefficient & smaller value is the Limiting Reactant.

           CaCO₃ => (2.15/3) = 0.72  and    AlF₃ => (3.83/2) = 1.92

The value for CaCO₃ (=0.72)          <        the value for AlF₃ (=1.92)  => CaCO₃ is the Limiting Reactant

Note: When working problem using mole ratios, use mole value for Limiting Reactant and not the value used to determine Limiting Reactant, in this case 2.15 moles CaCO₃.

grams CaF₂ Produced:

moles CaF₂ produced = 3/3(2.15) moles CaF₂ = 2.15 moles CaF₂

grams CaF₂ produced = 2.15 moles CaF₂ x 78 g CaF₂ / mole CaF₂

= 167.7 grams CaF₂

grams Al₂(CO₃)₃ produced:

moles Al₂(CO₃)₃ produced = 1/3(2.15) moles Al₂(CO₃)₃ = 0.72 mole Al₂(CO₃)₃

grams Al₂(CO₃)₃ = 0.72  mole Al₂(CO₃)₃  x  234 g Al₂(CO₃)₃ / mole Al₂(CO₃)₃ = 167.7 grams  Al₂(CO₃)₃  

it takes on an electronical energy