if you have 66.6g NH3, how many grams of F2 are required for a complete reaction?
please show work!!!!!
To find out how many grams of F2 are required for a complete reaction, we'll use stoichiometry. The balanced chemical equation for the reaction of NH3 with F2 is:
2 NH3 + 3 F2 -> 2 NF3 + 3 H2
This means that 2 moles of NH3 react with 3 moles of F2 to produce 2 moles of NF3 and 3 moles of H2.
First, we need to convert the given mass of NH3 into moles:
moles of NH3 = mass of NH3 / molar mass of NH3
Molar mass of NH3 = 14.01 g/mol (N) + 3 * 1.01 g/mol (H) = 17.04 g/mol
moles of NH3 = 66.6 g / 17.04 g/mol = 3.91 moles
Now, we'll use the stoichiometry from the balanced equation to find out the moles of F2 required:
moles of F2 = moles of NH3 * (3 moles of F2 / 2 moles of NH3)
moles of F2 = 3.91 moles * (3/2) = 5.865 moles
Finally, we'll convert the moles of F2 into grams:
mass of F2 = moles of F2 * molar mass of F2
Molar mass of F2 = 2 * 19.00 g/mol (F) = 38.00 g/mol
mass of F2 = 5.865 moles * 38.00 g/mol = 222.87 g
Therefore, 222.87 grams of F2 are required for a complete reaction with 66.6 grams of NH3.
To determine the amount of F2 required for a complete reaction with 66.6g NH3, we need to first balance the chemical equation between NH3 and F2.
The balanced chemical equation for the reaction between NH3 and F2 is:
8 NH3 + 15 F2 -> 10 NF3 + 6 HF
From the balanced equation, we can see that the molar ratio between NH3 and F2 is 8:15. This means that for every 8 moles of NH3, we need 15 moles of F2.
To find the number of moles of NH3, we can use the molar mass of NH3. The molar mass of NH3 is calculated as:
(1 * 3) + (14 * 1) = 17 g/mol.
Therefore, the number of moles of NH3 is:
66.6g / 17 g/mol = 3.91 mol.
Since the molar ratio between NH3 and F2 is 8:15, we can find the number of moles of F2 by multiplying the number of moles of NH3 by the mole ratio:
3.91 mol NH3 * (15 mol F2 / 8 mol NH3) = 7.27 mol F2.
Finally, to find the mass of F2 required, we can multiply the number of moles of F2 by its molar mass. The molar mass of F2 is 2 * 19 = 38 g/mol. Therefore, the mass of F2 required is:
7.27 mol F2 * 38 g/mol = 276.26 g F2.
So, 276.26 grams of F2 are required for a complete reaction with 66.6 grams of NH3.
To determine the amount of F2 required for a complete reaction with 66.6 grams of NH3, we first need to write the balanced chemical equation for the reaction between NH3 and F2.
The balanced equation is:
2 NH3 + 3 F2 → 6 HF + N2
From the balanced equation, we can see that the stoichiometric ratio between F2 and NH3 is 3:2. This means that for every 3 moles of F2, 2 moles of NH3 are needed.
Step 1: Convert the mass of NH3 to moles.
To do this, we need to know the molar mass of NH3.
Molar mass of NH3 = (1 x atomic mass of N) + (3 x atomic mass of H)
= (1 x 14.01 g/mol) + (3 x 1.01 g/mol)
≈ 17.03 g/mol
Mass of NH3 = 66.6 g
Moles of NH3 = Mass of NH3 / Molar mass of NH3
= 66.6 g / 17.03 g/mol
≈ 3.91 moles
Step 2: Determine the moles of F2 needed for the reaction.
From the stoichiometric ratio, we know that 2 moles of NH3 react with 3 moles of F2.
Moles of F2 = (Moles of NH3 x Coefficient of F2) / Coefficient of NH3
= (3.91 moles NH3 x 3) / 2
= 5.87 moles
Step 3: Convert moles of F2 to grams.
To do this, we need to know the molar mass of F2.
Molar mass of F2 = (2 x atomic mass of F)
= (2 x 19.00 g/mol)
= 38.00 g/mol
Mass of F2 = Moles of F2 x Molar mass of F2
= 5.87 moles x 38.00 g/mol
≈ 222.94 g
Therefore, approximately 222.94 grams of F2 are required for the complete reaction with 66.6 grams of NH3.