Answer:
See explanation
Explanation:
2)CH4 + 2O2 -------> CO2 + 2H2O
96g of O2 = 96g/32g/mol = 3 moles of O2
If 1 mole of CH4 reacts with 2 moles of O2
x moles of CH4 reacts with 3 Moles of O2
x = 3/2 = 1.5 moles of CH4
Mass of CH4 = 1.5 moles * 16 = 24 g of CH4
3)C6H6 + HNO3 ---->C6H5NO2 + H2
Number of moles in 15 g of benzene = 15 g/ 78 g/mol = 0.19 moles
Number of moles of 18 g of C6H5NO2 = 18g/123 g/mol = 0.15 moles
From the reaction equation;
1 mole of C6H6 yields 1 mole of C6H5NO2
Hence 0.19 moles of C6H6 also yields 0.19 moles of C6H5NO2
% yield = actual yield/ theoretical yield * 100
actual yield = 0.15 moles
Theoretical yield = 0.19 moles
% yield = 0.15/0.19 * 100
% yield = 78.9%
4)The reaction equation is
KOH + CH3COOH ------> CH3COOK + H2O
number of moles of acetic acid reacted = 0.215g/60 g/mol = 3.58 * 10^-3 moles
Since the reaction is 1:1, 3.58 * 10^-3 moles moles of KOH is used.
But;
n = CV
n = number of moles
C = concentration of reactants
V = volume of solution
V = n/C
V= 3.58 * 10^-3 moles/0.225M = 0.016 L
5) 3AgNO3 + AlCl3 -----> Al(NO3) + 3AgCl
Number of moles in 0.325 g of AgCl = 0.325g/143 g/mol = 2.27 * 10^-3 moles
From the reaction equation;
1 mole of AlCl3 yields 3 moles of AgCl
x moles of AlCl3 yields 2.27 * 10^-3 moles of AgCl
x = 2.27 * 10^-3 moles/3
x = 7.6 * 10^-4 moles
From;
n = CV
C = n/V
C = 7.6 * 10^-4 moles * 1000/185.5
C= 4.1 * 10^-3 MolL-1
6)From;
C1V1 =C2V2
Where;
C1 = initial concentration
V1= initial volume
C2= final concentration
V2= final volume
100 * 12 = 4.75 * V2
V2 = 100 * 12/4.75
V2 = 252.6 ml
Volume of water added = 252.6 - 100 = 152.6 ml of pure water
