4. A 20.25 ml unknown HaSO, solution is titrated with 34.60 ml ofa 0.10M NaOH to

Question

4. A 20.25 ml unknown HaSO, solution is titrated with 34.60 ml ofa 0.10M NaOH to the equivalence point. Find the concentration of the HaSO4 solution. (6 pt) 5. Name or write the formula for the following acids or bases (2 pt each c.phospboric acid ????? a. HCIO2 b.NH OH 6. Desemine whether each reaction is a redox reaction and identifly the oxdidizing sgent and dhe reducing agent (3 pt) 7. Assign oxidation states to the bold italicized atom. (3 pt each) 8. At 1.0 atm pressure, air contains 21% O2 and 78% N. Calculate the mass of oxygen gas present in the mixture with a total pressure of 700 mm Hg at 298 "K in a 2500 ml container. (8 pt) 9. Ferric oxide reacts with carbon monoxide gas at high temperature and 2.0 atm pressure to produce pig the following reaction. How much iron can be produced in a 10.0 L container iron (Fe) according to at2.0 atmC?? pressure and 1250?? (8pt)

Explanation / Answer

Ans. #4. Balanced Reaction: H2SO4 + 2 NaOH ----------> Na2SO4 + H2O

At equivalence point, the total number of moles of H+ from acid is equal to total number of OH- from base.

That is-           x(M1V1), acid = y(M2V2), base             - equation 1

            Where, x = moles of H+ produced per mol acid = 2 for H2SO4

                        y = moles of OH- produced per mol base = 1 for NaOH

                        V and M are volume and molarity of respective solution.

# Putting the values in equation 1-

            2 x (M1 x 20.25 mL) = 1 x (0.10 M x 34.60 mL)

            Or, M1 = (0.10 M x 34.60 mL) / (2 x 20.25 mL)

            Hence, M1 = 0.085 M

Therefore, required molarity of H2SO4 solution = 0.085 M

#5. #b. NH4OH = Ammonium hydroxide

#6. Oxidizing agent: A chemical species that oxidizes other chemical spices by getting itself reduced is called an oxidant agent. Reduction is the decrease in oxidation number of a chemical species.

The oxidation number of Cu decreases from +2 in Cu2+ to 0 in Cu(s). So, Cu2+ is getting reduced during the process. So, Cu2+ is the oxidant agent.

Reducing agent: A chemical species that reduces other chemical spices by getting itself oxidized is called a reducing agent. Oxidation is the increase in oxidation number of a chemical species.

The oxidation number of Na increases from 0 in Na(s) to +1 in Na+. So, Na(s) is the reducing agent because it’s getting itself oxidized.

#7. #a. Note that ClO4 has a net charge on -1. The oxidation number of O is taken -2. The sum of oxidation number of constituent chemical species is equal to the net charge on the ion.

Let the oxidation number of Cl be X.

Now,

            (X x 1) + (-2 x 4) = -1

            Or, X = -1 + 8 = +7

Hence, oxidation number of Cl = +7

#b. Since no charge is indicated on the molecule, it’s neutral. The oxidation number of O is -2 ; the oxidation number of H is +1 The oxidation number of H is The sum of oxidation number of constituent chemical species is equal zero for a neutral molecule.

Let the oxidation number of P be X.

Now,

            (+1 x 3) + (X) + (-2 x 4)= 0

            Or, +3 + X – 8 = 0

            Hence, X = +5

Hence, oxidation number of P = +5

#8. Step 1: Using Ideal gas equation:       PV = nRT     

            Where, P = pressure in atm

            V = volume in L                   

            n = number of moles

            R = universal gas constant= 0.0821 atm L mol-1K-1

            T = absolute temperature (in K) = (0C + 273.15) K

Given, Pressure P = 700 mmHg = (700 / 760) atm = 0.9211 atm

            Volume, V = 2500 mL = 2.50 L

# Putting the values in ideal gas equation-

            0.9211 atm x 2.50 L = n x 0.0821 atm L mol-1K-1 x 298 K

            Or, n = 25.00 atm L / 24.4658 atm L mol-1

            Hence, n = 0.1022 mol

Therefore, total moles of both the gases in the container = 0.1022 mol

# Step 2: Following Avogadro’s Law, equal volume of all gases have equal number of moles, at constant pressure and temperature.

So,

            Moles of O2 = 21 % of 0.1022 mol = 0.021462 mol

Now,

            Mass of O2 = moles x MW = 0.021462 mol x 32.0 g mol-1 = 0.686784 g

#9. Step 1: Calculate moles of CO

# Putting the values in ideal gas equation-

            2.0 atm x 10.0 L = n x 0.0821 atm L mol-1K-1 x 298 K

            Or, n = 20.00 atm L / 125.0506 atm L mol-1

            Hence, n = 0.080 mol

Hence, moles of CO in the reaction vessel = 0.080 mol

# Step 2: Following stoichiometry, 3 mol CO produces 2 mol Fe.

So, assuming completion of reaction-

            Moles of Fe formed = (2 / 3) x Moles of CO = (2/ 3) x 0.080 mol = 0.053 mol

Now,

            Mass of Fe formed = Moles x MW = 0.053 mol x 55.847 g mol-1 = 2.96 g

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