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How many grams of the excess reagent are left over if 37.8 g of [tex]Cl_2[/tex] react with 39.4 g of [tex]NaF[/tex]?

\[ \text{Reaction: } Cl_2(g) + 2NaF(aq) \rightarrow 2NaCl(aq) + F_2(g) \]

Answer :

If 37.8 g of Cl2 react with 39.4 g of NaF, there will be no grams of the excess reagent left over.

The given reaction is: Cl2(g) + 2 NaF(aq) → 2 NaCl(aq) + F2(g)
To determine the grams of the excess reagent left over, we need to follow these steps:
1. Calculate the molar mass of Cl2 and NaF.
- The molar mass of Cl2 is 70.906 g/mol (35.453 g/mol x 2).
- The molar mass of NaF is 41.997 g/mol (22.990 g/mol + 18.998 g/mol).
2. Use the molar masses to convert the given masses into moles.
- Moles of Cl2 = 37.8 g / 70.906 g/mol = 0.532 mol
- Moles of NaF = 39.4 g / 41.997 g/mol = 0.939 mol

3. Determine the limiting reagent.
- To find the limiting reagent, we compare the moles of each reactant to the stoichiometric ratio in the balanced equation.
- The stoichiometric ratio is 1 mole of Cl2 to 2 moles of NaF.
- Since we have 0.532 mol of Cl2 and 0.939 mol of NaF, the limiting reagent is Cl2.
4. Calculate the moles of the excess reagent that reacted.
- Since Cl2 is the limiting reagent, it will be completely consumed in the reaction.
- Therefore, none of the Cl2 will be left over.
5. Calculate the grams of the excess reagent left over.
- Since none of the Cl2 is left over, the mass of the excess reagent is 0 grams.

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