what do limiting reagents have to do with...
5F2(g) + 2NH3(g) ==> N2F4(g) + 6HF(g)
a)If you have 66.6 g NH3, how many grams of F2 are required for complete reaction?
b)How many grams of NH3 are required to produce 4.65g HF?
c)How many grams of N2F4 can be produced from 225g F2?
In the given chemical equation:
5F2(g) + 2NH3(g) ==> N2F4(g) + 6HF(g)
The coefficients represent the stoichiometric ratio between the reactants and products. The limiting reagent is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
a) To find the grams of F2 required for complete reaction with 66.6 g NH3, we need to determine the limiting reagent by comparing the number of moles of each reactant.
First, calculate the molar mass of NH3:
Molar mass of NH3 = 14.01 g/mol (N) + 1.01 g/mol (H) x 3 = 17.04 g/mol
Next, convert grams of NH3 to moles:
66.6 g NH3 x (1 mol NH3/17.04 g NH3) = 3.90 mol NH3
Now, let's calculate the number of moles of F2 required using the stoichiometric ratio:
From the balanced equation, the ratio of F2 to NH3 is 5:2.
So, 2 moles of NH3 react with 5 moles of F2.
(3.90 mol NH3) x (5 mol F2/2 mol NH3) = 9.75 mol F2
Finally, convert moles of F2 to grams:
Molar mass of F2 = 19.00 g/mol
9.75 mol F2 x 19.00 g F2/mol = 185.25 g F2
Therefore, 185.25 grams of F2 are required for complete reaction with 66.6 grams of NH3.
b) To find the grams of NH3 required to produce 4.65 g HF, we will use the same approach.
First, calculate the molar mass of HF:
Molar mass of HF = 1.01 g/mol (H) + 19.00 g/mol (F) = 20.01 g/mol
Next, convert grams of HF to moles:
4.65 g HF x (1 mol HF/20.01 g HF) = 0.232 mol HF
Now, use the stoichiometric ratio to find the moles of NH3:
From the balanced equation, the ratio of NH3 to HF is 2:6.
So, 2 moles of NH3 react with 6 moles of HF.
(0.232 mol HF) x (2 mol NH3/6 mol HF) = 0.0773 mol NH3
Finally, convert moles of NH3 to grams:
0.0773 mol NH3 x 17.04 g NH3/mol = 1.32 g NH3
Therefore, 1.32 grams of NH3 are required to produce 4.65 grams of HF.
c) To determine the grams of N2F4 that can be produced from 225 g F2, we will follow a similar process.
First, calculate the molar mass of N2F4:
Molar mass of N2F4 = 28.02 g/mol (N) x 2 + 19.00 g/mol (F) x 4 = 92.02 g/mol
Next, convert grams of F2 to moles:
225 g F2 x (1 mol F2/19.00 g F2) = 11.8 mol F2
Now, use the stoichiometric ratio to find the moles of N2F4:
From the balanced equation, the ratio of F2 to N2F4 is 5:1.
So, 5 moles of F2 react with 1 mole of N2F4.
(11.8 mol F2) x (1 mol N2F4/5 mol F2) = 2.36 mol N2F4
Finally, convert moles of N2F4 to grams:
2.36 mol N2F4 x 92.02 g N2F4/mol = 217 g N2F4
Therefore, 217 grams of N2F4 can be produced from 225 grams of F2.
Limiting reagents are substances that are completely consumed in a chemical reaction, thereby limiting the amount of product that can be formed. To determine the limiting reagent, you need to compare the mole ratios of the reactants as given in the balanced chemical equation.
Let's break down each part of the question:
a) To find the grams of F2 required for complete reaction, you need to identify the limiting reagent first. In this case, we are given the mass of NH3.
Step 1: Convert the given mass of NH3 to moles using its molar mass.
NH3 molar mass = 17.03 g/mol
Moles of NH3 = (mass of NH3 given) / (molar mass of NH3)
Step 2: Use the balanced chemical equation to compare the mole ratio of NH3 to F2.
From the reaction: 5 moles of F2 react with 2 moles of NH3.
Step 3: Convert the moles of NH3 to moles of F2 using the mole ratio.
Moles of F2 = (moles of NH3) x (5 moles of F2 / 2 moles of NH3)
Step 4: Convert the moles of F2 to grams using its molar mass.
F2 molar mass = 38.00 g/mol
Mass of F2 = (moles of F2) x (molar mass of F2)
b) To find the grams of NH3 required to produce 4.65 g HF, we will follow a similar approach.
Step 1: Convert the given mass of HF to moles using its molar mass.
HF molar mass = 20.01 g/mol
Moles of HF = (mass of HF given) / (molar mass of HF)
Step 2: Use the balanced chemical equation to compare the mole ratio of NH3 to HF.
From the reaction: 2 moles of NH3 react with 6 moles of HF.
Step 3: Convert the moles of HF to moles of NH3 using the mole ratio.
Moles of NH3 = (moles of HF) x (2 moles of NH3 / 6 moles of HF)
Step 4: Convert the moles of NH3 to grams using its molar mass.
NH3 molar mass = 17.03 g/mol
Mass of NH3 = (moles of NH3) x (molar mass of NH3)
c) To find the grams of N2F4 that can be produced from 225 g F2, we will again use a similar approach.
Step 1: Convert the given mass of F2 to moles using its molar mass.
F2 molar mass = 38.00 g/mol
Moles of F2 = (mass of F2 given) / (molar mass of F2)
Step 2: Use the balanced chemical equation to compare the mole ratio of F2 to N2F4.
From the reaction: 5 moles of F2 react with 1 mole of N2F4.
Step 3: Convert the moles of F2 to moles of N2F4 using the mole ratio.
Moles of N2F4 = (moles of F2) x (1 mole of N2F4 / 5 moles of F2)
Step 4: Convert the moles of N2F4 to grams using its molar mass.
N2F4 molar mass = 104.00 g/mol
Mass of N2F4 = (moles of N2F4) x (molar mass of N2F4)
Following these steps will help you calculate the grams of the respective compounds as required in each part of the question.
Well, it has a lot to do with it since instructions for working a limiting reagent problem is just an expanded set of rules (some don't apply) when doing a simpler stoichiometry problem. I have been handy tonight with passing out answers I've given, especially for limiting reagent problems, because I got tired of typing the same thing over and over again (and for the same problem, yet). However, if you got caught in the crossfire, I apologize; but, you can use that set if you ignore the items that don't apply. I worked this problem for someone but here it is again.
a. Convert 66.6 g NH3 to moles. moles = grams/molar mass. Using the coefficients in the balanced equation, convert moles NH3 to moles F2. Now convert moles F2 to grams using g = moles x molar mass.
b. Convert 5.65 g HF to moles. Using the coefficients in the balanced equation, convert moles HF to moles NH3. Now convert moles NH3 to grams. (See the similarity?)
c. Convert 225 g F2 to moles. Using the coefficients in the balanced equation, convert moles F2 to moles N2F4. Convert to grams.