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3CaCl2 + 2AlF3 --> 3CaF2 + 2AlCl3 3CaCl2 + 2AlF3 --> 3CaF2 + 2AlCl3 How many grams of CaF2 will form from 36.5 grams of AlF3?

Sagot :

maacastrobr

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

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