Need help with my last question!
Given the following reaction:
2NH3 (g) + 5F2 (g) ---> N2F4 (g) + 6HF (g)
If 4.00 g of NH3 and 14.0 g of F2 are allowed to react, how many grams of HF can be produced?
grams/mol
N=14
H=1
F= 19
so
NH3 =17
F2 = 38
HF = 20
so
If I have 4 g of NH3, how many g of F2 do I need?
4/[2(17)] = x/[5(38)]
x = 20*38/34 = 22.4 g F2
whoops, not enough F2
so use 14 of F2 and y of NH3
y/34 = 14/190
y = 2.5 grams of NH3
then
2.5/34 = HF/[6(20)]
HF = 8.82 grams HF
=======================
check that should be 14 grams of F2
14/[5(38)] = 8.8/[6(20)]
.074 = .073 close enough
To find the number of grams of HF produced in the given reaction, we first need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed, which will determine the maximum amount of product that can be formed.
To find the limiting reactant, we can use stoichiometry and convert the given masses of NH3 and F2 into the number of moles.
1. Calculate the number of moles of NH3:
moles of NH3 = mass of NH3 / molar mass of NH3
The molar mass of NH3 (ammonia) is:
Molar mass of N = 14.01 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of NH3 = (3 * Molar mass of H) + Molar mass of N
= (3 * 1.01 g/mol) + 14.01 g/mol
= 17.04 g/mol
Therefore, moles of NH3 = 4.00 g / 17.04 g/mol
2. Calculate the number of moles of F2:
moles of F2 = mass of F2 / molar mass of F2
The molar mass of F2 (fluorine gas) is:
Molar mass of F = 19.00 g/mol
Therefore, moles of F2 = 14.0 g / 19.00 g/mol
3. Now, we need to determine the limiting reactant by comparing the mole ratios between NH3 and F2 in the balanced equation. The balanced equation tells us that 2 moles of NH3 react with 5 moles of F2.
We can calculate the mole ratios for NH3 and F2:
mole ratio of NH3 to F2 = moles of NH3 / stoichiometric coefficient of NH3
= moles of NH3 / 2
mole ratio of F2 to NH3 = moles of F2 / stoichiometric coefficient of F2
= moles of F2 / 5
The reactant with the smaller mole ratio value is the limiting reactant.
4. Compare the mole ratios:
If mole ratio of NH3 to F2 < mole ratio of F2 to NH3, then NH3 is the limiting reactant.
If mole ratio of F2 to NH3 < mole ratio of NH3 to F2, then F2 is the limiting reactant.
Let's calculate the mole ratios:
mole ratio of NH3 to F2 = moles of NH3 / 2
= (moles of NH3) / 2
mole ratio of F2 to NH3 = moles of F2 / 5
= (moles of F2) / 5
Compare the mole ratios, and determine the limiting reactant.
5. Once the limiting reactant is identified, we can calculate the number of moles of HF that can be produced using stoichiometry. The balanced equation tells us that 2 moles of NH3 react to produce 6 moles of HF.
Calculate the number of moles of HF using the balanced equation and the moles of the limiting reactant.
6. Finally, convert the moles of HF to grams using the molar mass of HF.
Follow these steps to determine the number of grams of HF that can be produced.