The volume of 0.160 m Li2S solution required to completely react with 130 ml of 0.160 CO(NO3)2 is calculated as below
write the reacting equation
Co(NO3)2 + Li2S = 2LiNO3 + COS
find the moles of CO(NO3)2 = molarity x volume
= 130 ml x 0.160=20.8 moles
since the reacting moles between CO(NO3)2 to LiS is 1:1 the moles of LiS is also 20.8 moles
volume of Lis is therefore = moles of Lis/ molarity of LiS
= 20.8/0.160 = 130 Ml
Answer:
The equilibrium concentration of NO is 0.001335 M
Explanation:
Step 1: Data given
The equilibrium constant Kc is 0.0025 at 2127 °C
An equilibrium mixture contains 0.023M N2 and 0.031 M O2,
Step 2: The balanced equation
N2(g) + O2(g) ↔ 2NO(g)
Step 3: Concentration at the equilibrium
[N2] = 0.023 M
[O2] = 0.031 M
Kc = 0.0025 = [NO]² / [N2][O2]
Kc = 0.0025 = [NO]² / (0.023)(0.031)
[NO] = 0.001335 M
The equilibrium concentration of NO is 0.001335 M
Answer:
71.372 g or 0.7 moles
Explanation:
We are given;
- Moles of Aluminium is 1.40 mol
- Moles of Oxygen 1.35 mol
We are required to determine the theoretical yield of Aluminium oxide
The equation for the reaction between Aluminium and Oxygen is given by;
4Al(s) + 3O₂(g) → 2Al₂O₃(s)
From the equation 4 moles Al reacts with 3 moles of oxygen to yield 2 moles of Aluminium oxide.
Therefore;
1.4 moles of Al will require 1.05 moles (1.4 × 3/4) of oxygen
1.35 moles of Oxygen will require 1.8 moles (1.35 × 4/3) of Aluminium
Therefore, Aluminium is the rate limiting reagent in the reaction while Oxygen is the excess reactant.
4 moles of aluminium reacts to generate 2 moles aluminium oxide.
Therefore;
Mole ratio Al : Al₂O₃ is 4 : 2
Thus;
Moles of Al₂O₃ = Moles of Al × 0.5
= 1.4 moles × 0.5
= 0.7 moles
But; 1 mole of Al₂O₃ = 101.96 g/mol
Thus;
Theoretical mass of Al₂O₃ = 0.7 moles × 101.96 g/mol
= 71.372 g
