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Answer :
The reaction is as follows:
2 NaF + Cl₂ → 2 NaCl + F₂
Moles NaF = 39.4 g/ 42 g/mol = 0.938 mol
Moles Cl₂ = 37.8 g/ 70.9 g/mol = 0.533 mol
Find the theoretical amount of Cl₂ needed for the given NaF:
0.938 mol NaF (1 mol Cl₂/2 mol NaF) = 0.469 mol Cl₂
Thus, the excess reactant is Cl₂. 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 Cl₂
2 NaF + Cl₂ → 2 NaCl + F₂
Moles NaF = 39.4 g/ 42 g/mol = 0.938 mol
Moles Cl₂ = 37.8 g/ 70.9 g/mol = 0.533 mol
Find the theoretical amount of Cl₂ needed for the given NaF:
0.938 mol NaF (1 mol Cl₂/2 mol NaF) = 0.469 mol Cl₂
Thus, the excess reactant is Cl₂. 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 Cl₂
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