Aluminum oxide produced : = 79.152 g
<h3>Further explanation</h3>
Given
46.5g of Al
165.37g of MnO
Required
Aluminum oxide produced
Solution
Reaction
2 Al (s) + 3 MnO (s) → 3 Mn (s) + Al₂O₃ (s)
mol = mass : Ar
mol = 46.5 : 27
mol = 1.722
mol = 165.37 : 71
mol = 2.329
mol : coefficient ratio Al : MnO = 1.722/2 : 2.329/3 = 0.861 : 0.776
MnO as a limiting reactant(smaller ratio)
So mol Al₂O₃ based on MnO as a limiting reactant
From equation , mol Al₂O₃ :
= 1/3 x mol MnO
= 1/3 x 2.329
= 0.776
Mass Al₂O₃ (MW=102 g/mol) :
= 0.776 x 102
= 79.152 g
Answer:
1280J are required.
Explanation:
Heat of fusion is defined as the amount of heat required to change its state from liquid to solid at its melting point at constant pressure.
As heat of fusion of gold is 64J/g, there are required 64J to melt 1g of gold at its melting point. The energy required to melt 20g is:
20g * (64J/g) =
1280J are required
Answer:
The value of the heat capacity of the Calorimeter
= 54.4 
Explanation:
Given data
Heat added Q = 4.168 KJ = 4168 J
Mass of water
= 75.40 gm
Temperature change = ΔT = 35.82 - 24.58 = 11.24 ° c
From the given condition
Q =
ΔT +
ΔT
Put all the values in above equation we get
4168 = 75.70 × 4.18 × 11.24 +
× 11.24
611.37 =
× 11.24
= 54.4 
This is the value of the heat capacity of the Calorimeter.