If chlorine and fluorine react according to the reaction (to balance):
Cl2+ F2→ ClF3
And calculate the amount of product and of limiting reagent that remains, when 10 g of chlorine and
30 g of fluorine are reacted to give chlorine trifluoride.
First balance the equation. The product has a ratio of 1:3 for the atoms. So,
2Cl2 + 6Fl2 = 4ClF3
convert to moles of each reactant
The one with the lesser no. of moles needed will limit the reaction.
Then figure how many moles of product are produced, and go back to grams.
To calculate the amount of product and the limiting reagent that remains, we need to determine which reactant is the limiting reagent and then use stoichiometry to calculate the amount of product formed.
1. Calculate the molar mass of chlorine (Cl2):
Each chlorine atom has a molar mass of 35.5 g/mol, so the molar mass of Cl2 is 35.5 g/mol x 2 = 71 g/mol.
2. Calculate the number of moles of chlorine:
Number of moles of Cl2 = mass of Cl2 (in grams) / molar mass of Cl2
Number of moles of Cl2 = 10 g / 71 g/mol = 0.14 mol
3. Calculate the molar mass of fluorine (F2):
Each fluorine atom has a molar mass of 19 g/mol, so the molar mass of F2 is 19 g/mol x 2 = 38 g/mol.
4. Calculate the number of moles of fluorine:
Number of moles of F2 = mass of F2 (in grams) / molar mass of F2
Number of moles of F2 = 30 g / 38 g/mol = 0.79 mol
5. Use the balanced equation to determine the stoichiometric ratio between Cl2 and ClF3:
From the balanced equation: 1 mole of Cl2 reacts with 1 mole of ClF3.
6. Determine the limiting reagent:
To determine the limiting reagent, we compare the number of moles of each reactant to their stoichiometric ratio. The reactant that produces the lesser number of moles of the product is the limiting reagent.
In this case, both Cl2 and F2 have equal stoichiometric ratios with ClF3 (1:1).
The number of moles required to react completely is 0.14 mol of Cl2 or 0.14 mol of F2. Since we have more than enough F2 (0.79 mol), the limiting reagent is Cl2.
7. Calculate the moles of ClF3 formed:
Since Cl2 is the limiting reagent, the amount of ClF3 formed is equal to the number of moles of Cl2 used.
8. Calculate the amount of ClF3 formed in grams:
Mass of ClF3 formed = number of moles of ClF3 x molar mass of ClF3
Molar mass of ClF3 = 35.5 g/mol (molar mass of Cl) + 19 g/mol (molar mass of F) = 54.5 g/mol
Mass of ClF3 formed = 0.14 mol x 54.5 g/mol = 7.63 g
9. Calculate the amount of excess reagent remaining:
To find the amount of excess reagent (F2) remaining, we need to determine the moles of F2 used.
Number of moles of F2 used = number of moles of Cl2 x stoichiometric ratio (from the balanced equation)
Number of moles of F2 used = 0.14 mol x 1 = 0.14 mol
Amount of F2 remaining = total moles of F2 - moles of F2 used
Amount of F2 remaining = 0.79 mol - 0.14 mol = 0.65 mol
10. Calculate the mass of F2 remaining:
Mass of F2 remaining = amount of F2 remaining x molar mass of F2
Mass of F2 remaining = 0.65 mol x 38 g/mol = 24.7 g
Therefore, the amount of chlorine trifluoride formed is 7.63 grams, and the amount of excess fluorine (limiting reagent that remains) is 24.7 grams.