61mol * 4 = 244moles of P
<span>Classify is the answer hope this helps :)</span>
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
Molar mass Cu(OH)₂ = 97.561 g/mol
97.561 g Cu(OH)₂ --------------- 6.02x10²³ atoms
? g Cu(OH)₂ -------------------- 9.1x10²⁵ atoms
mass = 9.1x10²⁵ * 97.561 / 6.02x10²³
mass = 8.87x10²⁷ / 6.02x10²³
mass = 14734.2 g
hope this helps!
