The reaction between N₂ and F₂ gives Nitrogen trifluoride as the product. The balanced equation is;
N₂ + 3F₂ → 2NF₃
The stoichiometric ratio between N₂ and NF₃ is 1 : 2
Hence,
moles of N₂ / moles of F₂ = 1 / 2
moles of N₂ / 25 mol = 0.5
moles of N₂ = 0.5 x 25 mol = 12.5 mol
Hence N₂ moles needed = 12.5 mol
At STP (273 K and 1 atm) 1 mol of gas = 22.4 L
Hence needed N₂ volume = 22.4 L mol⁻¹ x 12.5 mol
= 280 L
Answer:
d properties
i took a test on this before
Q = ?
Cp = 0.397 J/ºC
Δt = 40.3 - 21.0<span> => 19.3</span><span> ºC</span>
m = 15.2 g
Q = m x Cp x Δt
Q = 15.2 x 0.397 x 19.3
Q ≈ 116.46 J
<span>hope this helps! </span>