If 2.34 moles of Mg react with 3.56 moles of l2 and 1.76 moles of Mgl2 form, what is the percent yield?
Answer:
The answer to your question is: V = 6.93 L
Explanation:
Data
N₂ = 5.6 g
Volume of NH₃ = ?
14 g of N ---------------- 1 mol
5.6 g ----------------------- x
x = (5.6 x 1) / 14 = 0.4 mol of N
Reaction
N₂ + 3H₂ ⇒ 2NH₃
1 mol of N₂ ---------------- 2 moles of NH₃
0.4 mol of N₂ -------------- x
x = (0.4 x 2) / 1
x = 0.8 mol of NH₃
Formula
PV = nRT
P = 5200 torr = 6.84 atm
V = ?
n = 0.8
R = 0.082 atm L/ mol °K
T = 450°C = 723°K
Substitution
V = (0.8)(0.082)(723) / 6.84
V = 6.93 L
Answer:
<u><em>See attachment for explanations.</em></u>
Explanation:
Answer:
96.5 g/ml
Explanation:
If 5g is 19.3 then 25g is 19.3x5 which is 96.5 g/ml
