5F2(g) + 2NH3(g) ==> N2F4(g) + 6HF(g) If you have 225 g F2, how many grams of N2F4 will be produced
The reaction of fluorine with ammonia produces dinitrogen tetrafluoride and hydrogen fluoride.
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 4.65 g HF?
C. How many grams of N2F4 can be produced from 225 g F2?
A. 372 g F2
B. 1.32 g NH3
C. 123 g N2F4
Tt
To determine the number of grams of N2F4 produced, we need to use stoichiometry and the given molar ratios from the balanced equation:
5F2(g) + 2NH3(g) ==> N2F4(g) + 6HF(g)
First, we need to calculate the number of moles of F2 using its molar mass. The molar mass of F2 is 2 x 19.00 g/mol = 38.00 g/mol.
Number of moles of F2 = Mass of F2 / Molar mass of F2
= 225 g / 38.00 g/mol
= 5.9211 mol (rounded to 4 decimal places)
According to the balanced equation, the stoichiometric ratio between F2 and N2F4 is 5:1. Therefore, for every 5 moles of F2 used, 1 mole of N2F4 is produced.
Since 5 moles of F2 produce 1 mole of N2F4, we can determine the number of moles of N2F4 as follows:
Number of moles of N2F4 = (Number of moles of F2) / (Stoichiometric ratio of F2 to N2F4)
= 5.9211 mol / 5
= 1.1842 mol (rounded to 4 decimal places)
Finally, we can calculate the mass of N2F4 using its molar mass, which is 92.00 g/mol:
Mass of N2F4 = Number of moles of N2F4 x Molar mass of N2F4
= 1.1842 mol x 92.00 g/mol
= 109.0408 g (rounded to 4 decimal places)
Therefore, approximately 109.04 grams of N2F4 will be produced when 225 grams of F2 reacts.
To calculate how many grams of N2F4 will be produced, we need to use the stoichiometry of the balanced chemical equation.
First, let's determine the molar mass of F2 (fluorine gas) and N2F4 (dinitrogen tetrafluoride). The molar mass of F2 is calculated as follows:
F2 molar mass = 2(Fluorine molar mass) = 2(19.00 g/mol) = 38.00 g/mol
Similarly, the molar mass of N2F4 is calculated as:
N2F4 molar mass = 2(Nitrogen molar mass) + 4(Fluorine molar mass)
= 2(14.01 g/mol) + 4(19.00 g/mol) = 92.02 g/mol
Next, we need to find the number of moles of F2 present in 225 g. To do this, we'll use the formula:
moles = mass / molar mass
moles of F2 = 225 g / 38.00 g/mol ≈ 5.92 mol
According to the balanced chemical equation, the molar ratio between F2 and N2F4 is 1:1. This means that for every 1 mol of F2, we produce 1 mol of N2F4.
Therefore, the number of moles of N2F4 produced will also be 5.92 mol.
Finally, we can determine the mass of N2F4 produced by multiplying the number of moles by the molar mass:
mass of N2F4 = moles of N2F4 × molar mass of N2F4
= 5.92 mol × 92.02 g/mol ≈ 546.06 g
Therefore, approximately 546.06 grams of N2F4 will be produced when starting with 225 grams of F2.