Answer:
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
