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1. Nitric oxide, NO, is made from the oxidation of NH3 as follows: 4NH3 + 5O2 4N

ID: 931241 • Letter: 1

Question

1. Nitric oxide, NO, is made from the oxidation of NH3 as follows: 4NH3 + 5O2 4NO + 6H2O

If 9.6-g of NH3 gives 12.0 g of NO, what is the percent yield of NO? (71.0 %)

2. How many moles of sulfate ions are there in a 0.547-L solution of 0.715 M Al2(SO4)3? (1.17 mol)

3. What minimum mass of cobalt (II) nitrate must be added to 60.0 mL of a 0.0890 M phosphate solution in order to completely precipitate all of the phosphate as solid cobalt (II) phosphate? (1.46 g)

2PO43–(aq) + 3Co(NO3)2(aq) Co3(PO4)2(s) + 6NO3–(aq)

4. The reaction of HCl with NaOH is represented by the equation

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

What volume of 0.668 M HCl is required to titrate 27.2 mL of 0.539 M NaOH?

(21.9 mL)

5. If 35.0 g H2O at 22.7 °C is combined with 65.0 g H2O at 87.5 °C, what is the final temperature of the mixture? The specific heat capacity of water is 4.184 J/g°C.

(64.8 oC)

6. CaO(s) reacts with water to form Ca(OH)2(aq). If 6.50 g CaO is combined with 99.70 g H2O in a coffee cup calorimeter, the temperature of the resulting solution increases from 21.7 °C to 43.1 °C. Calculate the enthalpy change for the reaction per mole of CaO. Assume that the specific heat capacity of the solution is 4.18 J/gK. Neglect the heat capacity of the calorimeter. (. -82.0 kJ/mol)

Hint: msolution=mwater+mCaO

7. What quantity, in moles, of hydrogen is consumed when 151.1 kJ of energy is evolved from the combustion of a mixture of H2(g) and O2(g)? (0.5286 mol)

H2(g) + O2(g) H2O(l); rH° = –285.8 kJ/mol-rxn

8. Calculate rH° for the combustion of ammonia,

4 NH3(g) + 7 O2(g) 4 NO2(g) + 6 H2O()

using standard molar enthalpies of formation. (–1398.8 kJ/mol)

molecule

fH° (kJ/mol)

NH3(g)

–45.9

NO2(g)

+33.1

H2O()

–285.8

9. Determine the enthalpy change for the decomposition of calcium carbonate

CaCO3(s) CaO(s) + CO2(g)

given the thermochemical equations below. (+179.0 kJ/mol-rxn)

Ca(OH)2(s) CaO(s) + H2O()

rH° = 65.2 kJ/mol-rxn

Ca(OH)2(s) + CO2(g) CaCO3(s) + H2O()

rH° = 113.8 kJ/mol-rxn

C(s) + O2(g) CO2(g)

rH° = 393.5 kJ/mol-rxn

2 Ca(s) + O2(g) 2 CaO(s)

rH° = 1270.2 kJ/mol-rxn

10. A red laser pointer emits light at a wavelength of 488 nm. If the laser emits 7.5 × 10–4 J of energy per second in the form of visible radiation, how many photons per second are emitted from the laser? (1.84 x 1015 photons/s)

11. Radio waves are very low energy forms of electromagnetic radiation. What is the energy of a photon of radio waves with a wavelength of 150. m? (1.33 x 10-27 J)

12. What is the binding energy of an electron in a photosensitive metal (in kJ/mol) if the longest wavelength of light that can eject electrons from the metal is 228.0 nm?

(525.0 kJ/mol)

Nitric oxide, NO ,is made from the oxidation of NH_3 as follow: 4NH_3+5O2 rightarrow 4NO+6H_2O If 9.6-g of NH3 gives 12.0 of NO, what is the percent yield of NO? (71.0%) How many moles of sulfate ions are there are there in a 0.547-L solution of 0.715 M A12(SO4)3? (1.17 mol) What minimum mass of cobalt (II) nitrate must be added to 60.0 mL of a 0.0890 M phosphate solution in order to complete precipitate all of the phosphate as solid cobalt (II) phosphate?(1.46g) 2PO43-(aq)+3Co(NO3)2(2q) rightarrow Co3(PO4)2(s)+6NO3-(aq) The reaction of HCl with NaOH is represented by equation HCl(aq)+NaOH(aq) rightarrow NaCl(aq)+H_2O(l) What volume of 0.668 M HCl is required to titrate 27.2 mL of 0.539 M NaOH? (21.9 mL) If 35.0 g H2O at 22.7 degree C is combined with 65.0 g H2o at 87.5 degreeC, what is the final temperature of the mixture? The specific heat capacity of water is 4.184 J/g. degree C. (64.8 oC) CaO(s) reacts with water to form Ca(OH)2(aq). If 6.50 g CaO is combined with 99.70 g H2O in a coffee cup calorimeter, the temperature of the resulting solution increases from 21.7 degree C to 43.1 degree C. Calculate the enthalpy change for the reaction per mole of Cao. Assume that the specific heat capacity of the solution is 4.18 J/g. K. Neglect the heat capacity of the calorimeter. (.-82.0 kJ/mol) What quantity, in moles , of hydrogen is consumed when 151.1 kJ of energy is evolved from the combustion of a mixture of H2(g) and O2(g)? (0.5286 mol) H2(g)+O2(g) rightarrow H2O(l); delta rH degree = -285.8 kJ/mol-rxn

Explanation / Answer

Answer – 1) We are given reaction - 4NH3 + 5O2 -----> 4NO + 6H2O

Mass of NH3 = 9.6 g , actual mass of NO = 12.0 g

First we need to calculate the moles of NH3

Moles of NH3 = 9.6 g / 17.0307 g.mol-1

                       = 0.564 moles

From the balanced equation

4 moles of NH3 = 4 moles of NO

So, 0.564 moles of NO = ?

= 0.564 moles of NO

Theoretical yield of NO = 0.564 moles * 30.006 g/mol

                                       = 16.91 g

We know percent yield = actual yield / theoretical yield * 100 %

                                     = 12.0 g / 16.91 g * 100 %

                                     = 71 %

2) We are given, molarity of Al2(SO4)3 = 0.715 M , volume = 0.547 L

First we need to calculate moles of Al2(SO4)3

Moles of Al2(SO4)3 = 0.715 M * 0.547 L

                              = 0.391 moles

We know,

1 mole of Al2(SO4)3 = 3 moles of SO42-

So, 0.391 moles of Al2(SO4)3 = ?

= 1.17 moles

Moles of sulfate ions are there in a 0.547-L solution of 0.715 M Al2(SO4)3 is 1.17 moles.

3) We are given, molarity of PO43- = 0.0890 M , volume = 60.0 mL = 0.060 L

Reaction – 2 PO43- (aq) + 3Co(NO3)2(aq) ------> Co3(PO4)2(s) + 6NO3- (aq)

Moles of PO43- = 0.0890 M * 0.060 L

                         = 0.00534 moles

From the balanced reaction

2 moles of PO43- = 3 moles of Co(NO3)2

So, 0.00534 moles of PO43- = ?

= 0.00801 moles

We know,

Mass of Co(NO3)2= 0.00801 moles * 182.94 g/mol

                               = 1.46 g

4) We are given, [NaOH] = 0.539 M , volume = 27.2 mL , volume of HCl = ?

[HCl] = 0.668 M

HCl(aq) + NaOH(aq) -----> NaCl(aq) + H2O(l)

Moles of NaOH = 0.539 M * 0.0272 L

                           = 0.0147 moles

We know mole ratio is 1:1

So moles of HCl = 0.0147 moles

So, volume of HCl = 0.0147 moles / 0.668 M

                               = 0.0219 L

                                = 21.9 mL