Fe2O3 + 2Al ---> Al2O3 + 2Fe
Mole ratio Fe2O3 : Al = 1:2
No. of moles of Fe2O3 = Mass/RMM = 250 / (55.8 * 2 + 16 * 3) = 1.56641604 moles
No. of moles of Al = 150/27 = 5.555555555 moles.
Mole ratio 1 : 2. 1.56641604 * 2 = 3.13283208 moles of Al, but you have 5.555555555 moles of Al. So Al is in excess. All of it won't react.
So take the Fe2O3 and Fe ratio to calculate the mass of iron metal that can be prepared.
RMM of Fe2O3 / Mass of Fe2O3 = RMM of 2Fe / Mass of Fe 159.6 / 250 = 111.6 / x x = 174.8 g of Fe
Explanation:
A Giant Gas Cloud. A star begins life as a large cloud of gas.
A Protostar Is a Baby Star.
The T-Tauri Phase.
Main Sequence Stars.
Expansion into Red Giant.
Fusion of Heavier Elements.
Supernovae and Planetary Nebulae.
5F2 +2NH3 = N2f4 + 6HF
moles = mass/molar mass
mass =molar mass x mole
a) moles of Nh3 = 58.5/17 = 3.44 moles
the mole ratio of F2:NH3 = 5:2
the moles of F2 =3.44x5/2=8.6 moles
mass= 8.6 x 38 = 326.8 grams of F2
B) moles of Hf = 3.89/20 = 0.1945 moles
mole ratio NH3:Hf = 2:6
moles of NH3 = 0.1945 x2/6 = 0.0648 moles
mass =0.0648 x17 = 1.102 grams of NH3
C) moles of f2 = 217/38 =5.711 moles
mole ratio N2F6:F2 = 5:1
moles of N2F4 =5.711 x1/5 =1.142 moles
mass = 1.142 x104 = 118.77 grams of N2F4
I am pretty sure the correct answer is B.
It makes the most sense to me.
