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Answer :
The reaction is as follows:
2 NaF + Clâ‚‚ → 2 NaCl + Fâ‚‚
Moles NaF = 39.4 g/ 42 g/mol = 0.938 mol
Moles Clâ‚‚ = 37.8 g/ 70.9 g/mol = 0.533 mol
Find the theoretical amount of Clâ‚‚ needed for the given NaF:
0.938 mol NaF (1 mol Clâ‚‚/2 mol NaF) = 0.469 mol Clâ‚‚
Thus, the excess reactant is Clâ‚‚. When all of the limiting reactant, NaF, is consumed, the amount of excess reactant left would be:
Excess Reagent = (0.533 mol - 0.469 mol) * 70.9 g/mol = 4.54 g Clâ‚‚
2 NaF + Clâ‚‚ → 2 NaCl + Fâ‚‚
Moles NaF = 39.4 g/ 42 g/mol = 0.938 mol
Moles Clâ‚‚ = 37.8 g/ 70.9 g/mol = 0.533 mol
Find the theoretical amount of Clâ‚‚ needed for the given NaF:
0.938 mol NaF (1 mol Clâ‚‚/2 mol NaF) = 0.469 mol Clâ‚‚
Thus, the excess reactant is Clâ‚‚. When all of the limiting reactant, NaF, is consumed, the amount of excess reactant left would be:
Excess Reagent = (0.533 mol - 0.469 mol) * 70.9 g/mol = 4.54 g Clâ‚‚
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