Nitrogen (N) has a molar mass of about 14.007 g/mol, so you have
100.0 g N₂ = (100.0 g) • (1/28.014 mol/g) ≈ 3.570 mol N₂
Hydrogen (H) has a molar mass of about 1.008 g/mol, so
100.0 g H₂ = (100.0 g) • (1/2.016 mol/g) ≈ 49.60 mol H₂
In the balanced reaction, 1 mole of N₂ reacts with 3 moles of H₂ to make 2 moles of NH₃. We have
3.570 • 3 ≈ 10.71
so the reaction would use up all 3.570 mol N₂ and 10.71 mol H₂ to produce about 7.139 mol NH₃.
(A) H₂ is the excess reactant. There is an excess of 49.60 - 10.71 ≈ 38.89 mol H₂.
(B) N₂ is the limiting reactant. All of the N₂ gets consumed in the reaction.
(C) NH₃ has a molar mass of (14.007 + 3 • 1.008) g/mol = 17.031 g/mol. This reaction would theoretically yield about 7.139 mol NH₃, or
7.139 mol NH₃ = (7.139 mol) • (17.031 g/mol) ≈ 121.6 g NH₃
Answer:
10%
Step-by-step explanation:
2937 ÷ 2670 = 1.1
1.1 - 1 = 0.1
Answer:
b=d
x=
bStep-by-step explanation:
-9u + 3u = 9 + 9
-6u = 18
u = -18/6
u = -3
Thus, u is equal to -3
