10.73 of fluorine is mixed with 24.73 grams of calcium.
a) Write balance calculate
b) What is the linuity reactant
c) What is the excess reactant
d) Calculate the theoretical yield of product formula
e) If 5.80 grams of calcium fluoride are produced in lab calculate the percent yield
PLEASE HELP ME
a) F2 + Ca = CaF2
b) To find the limiting reactant, we need to calculate the number of moles of reactants:
n(F2) = V(F2) / V(mol) = 10.73 l / 22.4 l/mol = 0.48 mol
n(Ca) = m(Ca) / Mr(Ca) = 24.73 g / 40 g/mol = 0.62 mol
From here, F2 is the limiting reactant.
c) Calcium is the excess reactant.
d) mtheor(CaF2) = n(F2) × Mr(CaF2) = 0.48 mol × 78 g/mol = 37.44 g
e) %w = (mexp / mtheor) × 100 % = (5.80 g / 37.44 g) × 100 % = 15.5 %
Answer: 15.5%
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