The reaction of fluorine with ammonia produces dinitrogen tetrafluoride and hydrogen fluoride.
5 F2(g) + 2 NH3(g) N2F4(g) + 6 HF(g)
(a) If you have 73.7 g NH3, how many grams of F2 are required for complete reaction?
b) How many grams of NH3, are required to produce 5.25 g HF?
(c) How many grams of N2F4 can be produced from 269 g F2?
These are three separate stoichiometry problems. Here is a sample stoichiometry problem (solved example). Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
I did follow those steps and they didn't work for the first problem.
Do you have the answer? If so I shall try it and see what I obtain. Post your work and I'll look for the error.
I worked it VERY fast with approximate molar masses and came up with an answer close to 412 grams F2 required.
To answer these questions, we need to use stoichiometry and mole ratio to convert between the given substances.
(a) To find the grams of F2 required for the reaction, we can follow these steps:
Step 1: Find the molar mass of NH3
NH3 (ammonia) consists of 1 nitrogen (N) atom and 3 hydrogen (H) atoms.
Molar mass of N = 14.01 g/mol
Molar mass of H = 1.008 g/mol
Molar mass of NH3 = (14.01 g/mol) + (3 * 1.008 g/mol) = 17.03 g/mol
Step 2: Convert grams of NH3 to moles
Given: 73.7 g NH3
Using the molar mass of NH3 (17.03 g/mol):
73.7 g NH3 * (1 mol NH3 / 17.03 g NH3) = 4.33 mol NH3
Step 3: Determine the mole ratio between NH3 and F2
From the balanced chemical equation, the mole ratio between NH3 and F2 is 2:5. This means that for every 2 moles of NH3, 5 moles of F2 are required.
Step 4: Calculate the moles of F2 required
4.33 mol NH3 * (5 mol F2 / 2 mol NH3) = 10.83 mol F2
Step 5: Convert moles of F2 to grams
Using the molar mass of F2 (38.00 g/mol):
10.83 mol F2 * (38.00 g F2 / 1 mol F2) = 411.39 g F2
Therefore, approximately 411.39 grams of F2 are required for complete reaction.
(b) To find the grams of NH3 required to produce a given amount of HF, we can follow these steps:
Step 1: Find the molar mass of HF
Molar mass of H = 1.008 g/mol
Molar mass of F = 18.99 g/mol
Molar mass of HF = (1.008 g/mol) + (18.99 g/mol) = 20.00 g/mol
Step 2: Convert grams of HF to moles
Given: 5.25 g HF
Using the molar mass of HF (20.00 g/mol):
5.25 g HF * (1 mol HF / 20.00 g HF) = 0.26 mol HF
Step 3: Determine the mole ratio between NH3 and HF
From the balanced chemical equation, the mole ratio between NH3 and HF is 2:6. This means that for every 2 moles of NH3, 6 moles of HF are produced.
Step 4: Calculate the moles of NH3
0.26 mol HF * (2 mol NH3 / 6 mol HF) = 0.0867 mol NH3
Step 5: Convert moles of NH3 to grams
Using the molar mass of NH3 (17.03 g/mol):
0.0867 mol NH3 * (17.03 g NH3 / 1 mol NH3) = 1.47 g NH3
Therefore, approximately 1.47 grams of NH3 are required to produce 5.25 g of HF.
(c) To find the grams of N2F4 that can be produced from a given amount of F2, we can follow these steps:
Step 1: Find the molar mass of F2
Molar mass of F2 = 2 * 18.99 g/mol = 37.98 g/mol
Step 2: Convert grams of F2 to moles
Given: 269 g F2
269 g F2 * (1 mol F2 / 37.98 g F2) = 7.08 mol F2
Step 3: Determine the mole ratio between F2 and N2F4
From the balanced chemical equation, the mole ratio between F2 and N2F4 is 5:1. This means that for every 5 moles of F2, 1 mole of N2F4 is produced.
Step 4: Calculate the moles of N2F4
7.08 mol F2 * (1 mol N2F4 / 5 mol F2) = 1.416 mol N2F4
Step 5: Convert moles of N2F4 to grams
Using the molar mass of N2F4 (104.01 g/mol):
1.416 mol N2F4 * (104.01 g N2F4 / 1 mol N2F4) = 147.29 g N2F4
Therefore, approximately 147.29 grams of N2F4 can be produced from 269 grams of F2.