# Midterm 1

General Chemistry 2

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1) You want to make 650 mL of 0.150 M potassium hydrogen phosphate.

How many moles of potassium hydrogen phosphate do you need?

2) You have in stock a solution that is 0.800M potassium hydrogen phosphate.

How much of this solution would you use to make your solution?

3) How can you check the molarity of your final potassium hydrogen phosphate solution?

Potassium hydrogen phosphate: K2HPO4

1) nK2HPO4 = MK2HPO4 x VK2HPO4

nK2HPO4 = 0.150 x 650 x 10-3

nK2HPO4 = 9.75 x 10-2 mol

2) solution 1: 0.800 M potassium hydrogen phosphate

solution 2: 0.150M potassium hydrogen phosphate

During a dilution: conservation of the number of moles

⇒ n1 = n2

⇒ M1 x V1 = M2 x V2
⇒ V1 =  =  = 1.22 x 102 mL

3) Titration with an acidic solution with a known concentration (for example 1 M HCl solution)

Consider the single replacement reaction of zinc metal with hydrochloric acid.

1) Write the chemical equation of this reaction knowing that one mole of H2 is formed from one mole of Zn.

2) Which atom gains electrons, which one loses electrons?

3) What is the reducing agent?

4) Write the net ionic equation for this reaction.

1) Zn (s) + 2 HCl (aq) → ZnCl2 (aq) + H2 (g)

2) Oxidation state: Zn = 0; Zn in ZnCl2 = +2

Oxidation state: Cl in HCl = -1; Cl in ZnCl2 = -1

Oxidation state: H in HCl = +1; H in H2 = 0

⇒ Zn loses electrons while H gains electrons

3) The oxidation state of H decreases: H is reduced.

Zn is the reducing agent.

4) Zn (s) + 2 H+ (aq) + 2 Cl- (aq) → Zn2+ (aq) + 2 Cl- (aq) + H2 (g)

Net ionic equation for this reaction: Zn (s) + 2 H+ (aq) → Zn2+ (aq) + H2 (g)

You have a solution that is 12.0% sucrose (C12H22O11) by weight (w/w) and you want 35.1 mmol of sucrose.

How many grams of this solution do you need?

nsucrose = $\frac{{\mathrm{m}}_{\mathrm{sucrose}}}{{\mathrm{M}}_{\mathrm{sucrose}}}$       with Msucrose = 12 MC + 22 MH + 11 MO = 12 x 12.0 + 22 x 1.01 + 11 x 16.0 = 342 g.mol-1

12.0% sucrose by weight (w/w): msucrose = $\frac{12}{100}$ x msolution

Mass of this solution needed:

msolution = $\frac{100}{12}$ x msucrose

msolution = $\frac{100}{12}$ x nsucrose x Msucrose

msolution = $\frac{100}{12}$ x 35.1 x 10-3 x 342

msolution = 1.00 x 102 g

We have 10.0 g of sucrose (C12H22O11) and we want to study the combustion of this compound.

1) How many molecules do you have in 10.0 g of sucrose?

2) What is the chemical equation for the combustion of sucrose?

3) We have 12.0 g of oxygen. What is the limiting reagent of this reaction?

1) n = $\frac{\mathrm{N}}{{\mathrm{N}}_{\mathrm{A}}}$   and    n = $\frac{\mathrm{m}}{\mathrm{M}}$   with   M = 12 MC + 22 MH + 11 MO = 12 x 12.0 + 22 x 1.01 + 11 x 16.0 = 342 g.mol-1

⇒ N = n x NA = $\frac{\mathrm{m}}{\mathrm{M}}$ x NA

⇒ N = $\frac{10.0}{342}$ x 6.022 x 1023 = 1.76 x 1022 molecules of sucrose

2) C12H22O11 (l) + 12 O2 (g) → 12 CO2 (g) + 11 H2O (l)

3) nsucrose = $\frac{{\mathrm{m}}_{\mathrm{sucrose}}}{{\mathrm{M}}_{\mathrm{sucrose}}}$ = $\frac{10.0}{342}$ = 2.92 x 10-2 mol

Moles of O2 needed to completely react with 10.0 g of sucrose:

12 moles of O2 react with 1 mole of sucrose

⇒ nO2 needed = 12 x nsucrose = 3.51 x 10-1 mol

nO2 = $\frac{{\mathrm{m}}_{\mathrm{O}2}}{{\mathrm{M}}_{\mathrm{O}2}}$ = $\frac{12.0}{32}$ = 3.75 x 10-1 mol > nO2 needed

The sucrose is the limiting reagent of this reaction

We have a 0.145 M solution of iron(II) bromide (MFeBr2 = 215.65 g.mol-1).

How much of this solution do we need to get 4.23 g of solute?

Iron(II) bromide: FeBr2

nFeBr2 = MFeBr2 x VFeBr2   and   nFeBr2 = $\frac{{\mathrm{m}}_{\mathrm{FeBr}2}}{{\mathrm{M}}_{\mathrm{FeBr}2}}$

⇒ MFeBr2 x VFeBr2 = $\frac{{\mathrm{m}}_{\mathrm{FeBr}2}}{{\mathrm{M}}_{\mathrm{FeBr}2}}$

⇒ VFeBr2 (in L) =  x MFeBr2 (in mol.L-1)

⇒ VFeBr2 (in L) =  = 1.35 x 10-1 L

What volume of a 0.150 M solution of the strong base KOH must be added to completely neutralize 200.0 mL of a 0.120 M solution of H2SO4?

Equation: 2 KOH (aq) + H2SO4 (aq) → K2SO4 (aq) + 2 H2O (l)

2 mole of KOH is required to neutralized 1 mole of H2SO4

⇒ when the endpoint of the neutralization is reached: nKOH = 2 x nH2SO4

⇒ MKOH x VKOH = 2 x MH2SO4 x VH2SO4

⇒ VKOH =

⇒ VKOH =  = 320 mL