Answer:
32.1 g
Explanation:
3CaCl₂ + 2AlF₃ → 3CaF₂ + 2AlCl₃
First we convert 36.5 g of AlF₃ into moles, using its molar mass:
36.5 g ÷ 133.34 g/mol = 0.274 mol AlF₃
Then we convert 0.274 moles of AlF₃ into moles of CaF₂, using the stoichiometric coefficients of the reaction:
0.274 mol AlF₃ * [tex]\frac{3molCaF_2}{2molAlF_3}[/tex] = 0.411 mol CaF₂
Finally we convert 0.411 moles of CaF₂ into grams, using its molar mass:
0.411 mol * 78.07 g/mol = 32.1 g
