How much nitrogen trifluoride (NF3) can be produced from 7 grams of nitrogen (N2)?

To calculate the mass of nitrogen trifluoride (NF₃) produced from 7 grams of nitrogen (N₂), we start with the balanced chemical equation:

N₂ + 3F₂ → 2NF₃

This equation tells us that 1 mole of nitrogen reacts with 3 moles of fluorine to produce 2 moles of nitrogen trifluoride.

First, we need to convert the mass of nitrogen to moles. The molar mass of nitrogen (N₂) is approximately 28 grams/mol (since the atomic mass of N is about 14 g/mol).

To find the number of moles of nitrogen in 7 grams, we use the formula:

moles = mass (g) / molar mass (g/mol)

Using the values:

moles of N₂ = 7 g / 28 g/mol = 0.25 moles

Now, according to the stoichiometry of the reaction, 1 mole of N₂ produces 2 moles of NF₃. Therefore, 0.25 moles of N₂ will produce:

0.25 moles N₂ × 2 moles NF₃ / 1 mole N₂ = 0.5 moles NF₃

Next, we will find the mass of nitrogen trifluoride produced. The molar mass of NF₃ is calculated as follows:

1 nitrogen atom = 14 g/mol and 3 fluorine atoms = 3 × 19 g/mol = 57 g/mol.

Thus, the molar mass of NF₃ = 14 g/mol + 57 g/mol = 71 g/mol.

Now, we can calculate the mass of NF₃ produced by:

mass = moles × molar mass

mass of NF₃ = 0.5 moles × 71 g/mol = 35.5 grams

In conclusion, from 7 grams of nitrogen, we can produce 35.5 grams of nitrogen trifluoride (NF₃).