Question

2.

and chemical equation
k. Molarity, normality, molality and formality
What mass of CH4 in grams, is required to react with 96.0 grams of O2?(2points)
S. A 15.6 g sample of benzene is mixed with excess nitric acid. We isolate 18.0 g of
C6H5NO2. What is the percent yield of C6H5NO2 in this reaction? (2points)
4. Calculate the volume of a 0.225 M solution of potassium hydroxide, KOH,
required to react with 0.215 g of acetic acid, CH3COOH? (2points)
5. An excess of AgNO3 reacts with 185.5 mL of an AlCl3 solution to give 0.325 g
of AgCl. What is the concentration, in moles per liter, of the AlCl3 solution?
(2 points)
6. Calculate the volume of pure water that used to dilute 100. mL of 12 M NaOH to
4.75 M. (2points)
3
2 16
&​

Answer 1

See explanation

Step-by-step 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