Answer:
a) 1.61 mol
b) Al is limiting reactant
c) HBr is in excess
Explanation:
Given data:
Moles of Al = 3.22 mol
Moles of HBr = 4.96 mol
Moles of H₂ formed = ?
What is limiting reactant =
What is excess reactant = ?
Solution:
Chemical equation:
2Al + 2HBr → 2AlBr + H₂
Now we will compare the moles:
Al : H₂
2 : 1
3.22 : 1/2×3.22 = 1.61 mol
HBr : H₂
2 : 1
4.96 : 1/2×4.96 = 2.48 mol
The number of moles of H₂ produced by Al are less it will be limiting reactant while HBr is present in excess.
Moles of H₂ :
Number of moles of H₂ = 1.61 mol
Answer:
flourine(1681),carbon(1086),lithium(520) and potassium(419)
as we can see that the ionization energy of flourine is the highest than carbon than lithium and than potassium
Explanation:
i hope it will help you
Best Answer:<span> </span><span>LiCl
Cl and Li
glucose
BeF2
last question answer 3 strong electrostatic requires a lot of energy
does this help though??</span>
The azimuthal quantum number (l) determines its orbital angular momentum and describes the shape of the orbital.
s-orbitals (for example 1s, 2s) are spherically symmetric around the nucleus of the atom.
p-orbitals are dumb-bell shaped. l = 0,1...n-1, when l = 1, that is p subshell.
d-orbitals are butterfly shaped.
Answer:
Amount of excess Carbon (ii) oxide left over = 23.75 g
Explanation:
Equation of the reaction: Fe₂O₃ + 3CO ----> 2Fe + 3CO₂
Molar mass of Fe₂O₃ = 160 g/mol;
Molar mass of Carbon (ii) oxide = 28 g/mol
From the equation of reaction, 1 mole of Fe₂O₃ reacts with 3 moles of carbon (ii) oxide; i.e. 160 g of iron (iii) oxide reacts with 84 g (3 * 28 g) of carbon (ii) oxide
450 g of Fe₂O₃ will react with 450 * 84/180) g of carbon (ii) oxide = 236..25 g of carbon (ii) oxide
Therefore the excess reactant is carbon (ii) oxide.
Amount of excess Carbon (ii) oxide left over = 260 - 236.25
Amount of excess Carbon (ii) oxide left over = 23.75 g
