The reaction of fluorine with ammonia produces dinitrogen tetra fluoride and hydrogen fluorida.
5F2(g)+2NH(g)---> N2F4(g)+6HF(g)
A. If you have 66.6gNH3, how many grams of F2 are required for a complete reaction?
B. how many grams of NH3 are required to produce 4.65g HFCS?
C. How many grams of N2F4 can be produced from 225g F2?
mols NH3 = grams/molar mass
A. Convert mols NH3 to mols F2 using the coefficients the balanced equation.
?mols NH3 x (5 mols F2/2 mol NH3) = ? mols F2. Then convert mols to grams. g F2 = mols F2 x molar mass F2.
B. and C. are done the same way.
To answer these questions, we need to use stoichiometry, which is the quantitative relationship between the amounts of reactants and products in a balanced chemical equation. Here's how we can solve each question:
A. To find the number of grams of F2 required, we need to use the molar ratio between NH3 and F2 from the balanced equation. The molar ratio of F2 to NH3 is 5:2. Here's the step-by-step solution:
Step 1: Calculate the molar mass of NH3.
NH3 = 14.01 g/mol (atomic mass of nitrogen) + 3 * 1.01 g/mol (atomic mass of hydrogen)
NH3 = 17.03 g/mol
Step 2: Convert grams of NH3 to moles using the molar mass.
66.6 g NH3 * (1 mol NH3 / 17.03 g NH3) = 3.91 mol NH3
Step 3: Use the molar ratio to calculate the moles of F2 required.
5 mol F2 / 2 mol NH3 = x mol F2 / 3.91 mol NH3
x = (5 mol F2 / 2 mol NH3) * 3.91 mol NH3
x ≈ 9.77 mol F2
Step 4: Convert moles of F2 to grams using the molar mass of F2.
9.77 mol F2 * (38.00 g F2 / 1 mol F2) = 371.26 g F2
Therefore, approximately 371.26 grams of F2 would be required for a complete reaction.
B. We can use a similar approach to solve this question. The molar ratio of NH3 to HF is 2:6 or simplified to 1:3. Here's the solution:
Step 1: Calculate the molar mass of HF.
HF = 1.01 g/mol (atomic mass of hydrogen) + 18.998 g/mol (atomic mass of fluorine)
HF = 20.00 g/mol
Step 2: Convert grams of HFCS to moles using the molar mass.
4.65 g HF * (1 mol HF / 20.00 g HF) = 0.233 mol HF
Step 3: Use the molar ratio to calculate the moles of NH3 required.
1 mol NH3 / 3 mol HF = x mol NH3 / 0.233 mol HF
x = (1 mol NH3 / 3 mol HF) * 0.233 mol HF
x ≈ 0.078 mol NH3
Step 4: Convert moles of NH3 to grams using the molar mass of NH3.
0.078 mol NH3 * (17.03 g NH3 / 1 mol NH3) = 1.33 g NH3
Therefore, approximately 1.33 grams of NH3 would be required to produce 4.65 grams of HF.
C. To determine the grams of N2F4 produced, we can use the molar ratio between F2 and N2F4 from the balanced equation. The molar ratio of F2 to N2F4 is 5:1. Here's the solution:
Step 1: Calculate the molar mass of F2.
F2 = 2 * 18.998 g/mol (atomic mass of fluorine)
F2 = 37.996 g/mol
Step 2: Convert grams of F2 to moles using the molar mass.
225 g F2 * (1 mol F2 / 37.996 g F2) = 5.92 mol F2
Step 3: Use the molar ratio to calculate the moles of N2F4.
5 mol F2 / 1 mol N2F4 = 5.92 mol F2 / x mol N2F4
x = (5 mol F2 / 1 mol N2F4) * 5.92 mol F2
x ≈ 29.6 mol N2F4
Step 4: Convert moles of N2F4 to grams using the molar mass of N2F4.
29.6 mol N2F4 * (92.011 g N2F4 / 1 mol N2F4) = 2,724 g N2F4
Therefore, approximately 2,724 grams of N2F4 can be produced from 225 grams of F2.