Answer:
Yes, yield.
Explanation:
N2(g) + 3 H2(g) → 2 NH3 (g) balanced equation
First, find limiting reactant:
Moles H2 = 1.83 g x 1 mole/2 g = 0.915 moles H2
Moles N2 = 9.84 g N2 x 1 mole/28 g = 0.351 moles N2
The mole ratio of H2: N2 is 3:1, so H2 is limiting (0.915 is less than 3 x 0.351)
Theoretical yield of NH3 = 0.915 mol H2 x 2 mol NH3/3 mol H2 = 0.61 moles NH3
Calcium carbon oxygen, calcium chlorine , carbon hydrogen oxygen, hydrogen chlorine, calcium oxygen hydrogen. pretty sure they’re right
<span> 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)
from the reaction 2 mol 4 mol
from the problem 5.4 mol 10.8 mol
M(CO2) = 12.0 +2*16.0 = 44.0 g/mol
10.8 mol CO2 * 44.0 g CO2/1 mol CO2 = 475.2 g CO2 </span>≈480 = 4.8 * 10² g
Answer is C. 4.8*10² g.
Answer:
Na₁₁ = 1s² 2s² 2p⁶ 3s¹
Explanation:
Sodium is present in group 1.
It is alkali metal.
It has one valence electron.
The atomic number of sodium is 11.
Its atomic mass is 23 amu.
The longhand notation of electronic configuration of sodium can be written as,
Na₁₁ = 1s² 2s² 2p⁶ 3s¹
The electronic configuration in shorthand notation( noble gas) would be written as,
Na₁₁ = [Ne] 3s¹
Sodium loses its one valence electron to complete the octet and get stable thus form +1 cation.
It react with halogen and form salt. Such as sodium chloride.
2Na + Cl₂ → 2NaCl
