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?
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http://www.jiskha.com/display.cgi?id=1270494161
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a) To calculate the amount of F2 required, we need to use the stoichiometry of the balanced equation.
Looking at the balanced equation:
5F2(g) + 2NH3(g) → N2F4(g) + 6HF(g)
We can see that the molar ratio between F2 and NH3 is 5:2. This means that for every 5 moles of F2, we need 2 moles of NH3.
1) Convert grams of NH3 to moles:
Molar mass of NH3 = 14.01 g/mol + 3(1.01 g/mol) = 17.03 g/mol
moles of NH3 = mass of NH3 / molar mass of NH3
= 66.6 g / 17.03 g/mol
≈ 3.91 moles NH3
2) Use the mole ratio to determine the moles of F2 required:
moles of F2 = (moles of NH3) x (5 moles of F2 / 2 moles of NH3)
= 3.91 moles x (5/2)
≈ 9.77 moles F2
3) Convert moles of F2 to grams:
Molar mass of F2 = 19.00 g/mol
grams of F2 = moles of F2 x molar mass of F2
≈ 9.77 moles x 19.00 g/mol
≈ 185.63 g
Therefore, approximately 185.63 grams of F2 is required for complete reaction.
b) To calculate the amount of NH3 required to produce HF, we use the stoichiometry of the balanced equation.
1) Convert grams of HF to moles:
Molar mass of HF = 1.01 g/mol + 19.00 g/mol = 20.01 g/mol
moles of HF = mass of HF / molar mass of HF
= 4.65 g / 20.01 g/mol
≈ 0.232 moles HF
2) Use the mole ratio to determine the moles of NH3 required:
moles of NH3 = (moles of HF) x (2 moles of NH3 / 6 moles of HF)
= 0.232 moles x (2/6)
≈ 0.0773 moles NH3
3) Convert moles of NH3 to grams:
grams of NH3 = moles of NH3 x molar mass of NH3
≈ 0.0773 moles x 17.03 g/mol
≈ 1.32 g
Therefore, approximately 1.32 grams of NH3 are required to produce 4.65 grams of HF.
c) To calculate the amount of N2F4 that can be produced from F2, we use the stoichiometry of the balanced equation.
1) Convert grams of F2 to moles:
Molar mass of F2 = 19.00 g/mol
moles of F2 = mass of F2 / molar mass of F2
= 225 g / 19.00 g/mol
≈ 11.84 moles F2
2) Use the mole ratio to determine the moles of N2F4 produced:
moles of N2F4 = (moles of F2) x (1 mole of N2F4 / 5 moles of F2)
= 11.84 moles x (1/5)
≈ 2.37 moles N2F4
3) Convert moles of N2F4 to grams:
Molar mass of N2F4 = 28.01 g/mol + 2(19.00 g/mol) = 90.01 g/mol
grams of N2F4 = moles of N2F4 x molar mass of N2F4
≈ 2.37 moles x 90.01 g/mol
≈ 213.11 g
Therefore, approximately 213.11 grams of N2F4 can be produced from 225 grams of F2.
To answer these questions, we need to use the given balanced chemical equation and convert the given quantities based on the stoichiometry of the reaction.
a) If you have 66.6 g NH3, we need to determine the amount of F2 required for the complete reaction.
Step 1: Write down the balanced chemical equation:
5F2(g) + 2NH3(g) → N2F4(g) + 6HF(g)
Step 2: Calculate the molar mass of NH3:
Molar mass of NH3 = 1(atomic mass of N) + 3(atomic mass of H)
= 1(14.01 g/mol) + 3(1.01 g/mol)
= 17.03 g/mol
Step 3: Convert the given mass of NH3 to moles:
66.6 g NH3 * (1 mol NH3 / 17.03 g NH3) = 3.91 mol NH3
Step 4: Use the stoichiometric ratio to determine the amount of F2:
From the balanced equation, we see that 2 moles of NH3 react with 5 moles of F2.
So, using the ratio of F2 to NH3, we have:
5 mol F2 / 2 mol NH3 * 3.91 mol NH3 = 9.77 mol F2
Step 5: Convert moles of F2 to grams:
Molar mass of F2 = 2(atomic mass of F)
= 2(19.00 g/mol)
= 38.00 g/mol
9.77 mol F2 * (38.00 g F2 / 1 mol F2) = 371.26 g F2
Therefore, you would require 371.26 grams of F2 for the complete reaction.
b) To find out how many grams of NH3 are required to produce 4.65 g of HF, we follow a similar procedure.
Step 1: Write down the balanced chemical equation:
5F2(g) + 2NH3(g) → N2F4(g) + 6HF(g)
Step 2: Calculate the molar mass of HF:
Molar mass of HF = 1(atomic mass of H) + 1(atomic mass of F)
= 1(1.01 g/mol) + 1(19.00 g/mol)
= 20.01 g/mol
Step 3: Convert the given mass of HF to moles:
4.65 g HF * (1 mol HF / 20.01 g HF) = 0.2325 mol HF
Step 4: Use the stoichiometric ratio to determine the amount of NH3:
From the balanced equation, we see that 6 moles of HF react with 2 moles of NH3.
So, using the ratio of NH3 to HF, we have:
2 mol NH3 / 6 mol HF * 0.2325 mol HF = 0.0775 mol NH3
Step 5: Convert the moles of NH3 to grams:
0.0775 mol NH3 * (17.03 g NH3 / 1 mol NH3) = 1.32 g NH3
Therefore, you would require 1.32 grams of NH3 to produce 4.65 g of HF.
c) To find out how many grams of N2F4 can be produced from 225 g of F2, we again use the balanced chemical equation and follow a similar procedure.
Step 1: Write down the balanced chemical equation:
5F2(g) + 2NH3(g) → N2F4(g) + 6HF(g)
Step 2: Convert the given mass of F2 to moles:
225 g F2 * (1 mol F2 / 38.00 g F2) = 5.92 mol F2
Step 3: Use the stoichiometric ratio to determine the amount of N2F4:
From the balanced equation, we see that 5 moles of F2 react with 1 mole of N2F4.
So, using the ratio of N2F4 to F2, we have:
1 mol N2F4 / 5 mol F2 * 5.92 mol F2 = 1.18 mol N2F4
Step 4: Convert the moles of N2F4 to grams:
Molar mass of N2F4 = 2(atomic mass of N) + 4(atomic mass of F)
= 2(14.01 g/mol) + 4(19.00 g/mol)
= 104.04 g/mol
1.18 mol N2F4 * (104.04 g N2F4 / 1 mol N2F4) = 122.65 g N2F4
Therefore, you can produce 122.65 grams of N2F4 from 225 grams of F2.