Asked by Tristan
if you have 66.6g NH3, how many grams of F2 are required for a complete reaction?
please show work!!!!!
please show work!!!!!
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Answered by
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To find out how many grams of F2 are required for a complete reaction, we'll use stoichiometry. The balanced chemical equation for the reaction of NH3 with F2 is:
2 NH3 + 3 F2 -> 2 NF3 + 3 H2
This means that 2 moles of NH3 react with 3 moles of F2 to produce 2 moles of NF3 and 3 moles of H2.
First, we need to convert the given mass of NH3 into moles:
moles of NH3 = mass of NH3 / molar mass of NH3
Molar mass of NH3 = 14.01 g/mol (N) + 3 * 1.01 g/mol (H) = 17.04 g/mol
moles of NH3 = 66.6 g / 17.04 g/mol = 3.91 moles
Now, we'll use the stoichiometry from the balanced equation to find out the moles of F2 required:
moles of F2 = moles of NH3 * (3 moles of F2 / 2 moles of NH3)
moles of F2 = 3.91 moles * (3/2) = 5.865 moles
Finally, we'll convert the moles of F2 into grams:
mass of F2 = moles of F2 * molar mass of F2
Molar mass of F2 = 2 * 19.00 g/mol (F) = 38.00 g/mol
mass of F2 = 5.865 moles * 38.00 g/mol = 222.87 g
Therefore, 222.87 grams of F2 are required for a complete reaction with 66.6 grams of NH3.
2 NH3 + 3 F2 -> 2 NF3 + 3 H2
This means that 2 moles of NH3 react with 3 moles of F2 to produce 2 moles of NF3 and 3 moles of H2.
First, we need to convert the given mass of NH3 into moles:
moles of NH3 = mass of NH3 / molar mass of NH3
Molar mass of NH3 = 14.01 g/mol (N) + 3 * 1.01 g/mol (H) = 17.04 g/mol
moles of NH3 = 66.6 g / 17.04 g/mol = 3.91 moles
Now, we'll use the stoichiometry from the balanced equation to find out the moles of F2 required:
moles of F2 = moles of NH3 * (3 moles of F2 / 2 moles of NH3)
moles of F2 = 3.91 moles * (3/2) = 5.865 moles
Finally, we'll convert the moles of F2 into grams:
mass of F2 = moles of F2 * molar mass of F2
Molar mass of F2 = 2 * 19.00 g/mol (F) = 38.00 g/mol
mass of F2 = 5.865 moles * 38.00 g/mol = 222.87 g
Therefore, 222.87 grams of F2 are required for a complete reaction with 66.6 grams of NH3.
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