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

Question

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

in progress 0
Thái Dương 3 months 2021-08-09T12:15:36+00:00 1 Answers 3 views 0

Answers ( )

    0
    2021-08-09T12:17:13+00:00

    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  

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )