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  

adding or remove a gaseous reactant or product changes the concentrations. if the concentration of reactant or product is increased, the system will shift away from the side in which concentration was increased (i.e. if the concentration of reactants is increased, the system will shift toward the products. if more products are added, the system will shift to form more reactants). conversely, if the concentration of reactant or product is decreased, the system will shift toward the side in which concentration was decreased (i.e. if reactants are removed, the system will shift to form more reactants. if the concentration of products is decreased, the equilibrium will shift toward the products).

explanation:

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