# Midterm 1

General Chemistry 2

Check the solutions on the questions and answer that you have decided correctly

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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 0.800*M* 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: K_{2}HPO_{4}

1) n_{K2HPO4} = *M*_{K2HPO4} x V_{K2HPO4}

n_{K2HPO4} = 0.150 x 650 x 10^{-3}

n_{K2HPO4 }= 9.75 x 10^{-2} mol

2) solution 1: 0.800 *M* potassium hydrogen phosphate

solution 2: 0.150*M* potassium hydrogen phosphate

During a dilution: conservation of the number of moles

⇒ n_{1} = n_{2}

⇒ *M*_{1 }x V_{1} = *M*_{2} x V_{2}

⇒ V_{1} = $\frac{{M}_{2}\times {\mathrm{V}}_{2}}{{M}_{1}}$ = $\frac{0.150\times 650}{0.800}$ = 1.22 x 10^{2} 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 H_{2} 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) → ZnCl_{2} (aq) + H_{2} (g)

2) Oxidation state: Zn = 0; Zn in ZnCl_{2} = +2

Oxidation state: Cl in HCl = -1; Cl in ZnCl_{2} = -1

Oxidation state: H in HCl = +1; H in H_{2} = 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) → Zn^{2+} (aq) + 2 Cl^{-} (aq) + H_{2} (g)

Net ionic equation for this reaction: Zn (s) + 2 H^{+} (aq) → Zn^{2+} (aq) + H_{2} (g)

You have a solution that is 12.0% sucrose (C_{12}H_{22}O_{11}) by weight (w/w) and you want 35.1 mmol of sucrose.

How many grams of this solution do you need?

n_{sucrose} = $\frac{{\mathrm{m}}_{\mathrm{sucrose}}}{{\mathrm{M}}_{\mathrm{sucrose}}}$ with M_{sucrose} = 12 M_{C} + 22 M_{H} + 11 M_{O} = 12 x 12.0 + 22 x 1.01 + 11 x 16.0 = 342 g.mol^{-1}

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

Mass of this solution needed:

m_{solution} = $\frac{100}{12}$ x m_{sucrose}

m_{solution} = $\frac{100}{12}$ x n_{sucrose} x M_{sucrose}

m_{solution} = $\frac{100}{12}$ x 35.1 x 10^{-3} x 342

m_{solution} = 1.00 x 10^{2} g

We have 10.0 g of sucrose (C_{12}H_{22}O_{11}) 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 M_{C} + 22 M_{H} + 11 M_{O} = 12 x 12.0 + 22 x 1.01 + 11 x 16.0 = 342 g.mol^{-1}

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

⇒ N = $\frac{10.0}{342}$ x 6.022 x 10^{23} = 1.76 x 10^{22} molecules of sucrose

2) C_{12}H_{22}O_{11} (l) + 12 O_{2} (g) → 12 CO_{2} (g) + 11 H_{2}O (l)

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

Moles of O_{2} needed to completely react with 10.0 g of sucrose:

12 moles of O_{2} react with 1 mole of sucrose

⇒ n_{O2 needed} = 12 x n_{sucrose} = 3.51 x 10^{-1} mol

n_{O2} = $\frac{{\mathrm{m}}_{\mathrm{O}2}}{{\mathrm{M}}_{\mathrm{O}2}}$ = $\frac{12.0}{32}$ = 3.75 x 10^{-1} mol > n_{O2 needed}

The sucrose is the limiting reagent of this reaction

We have a 0.145 *M* solution of iron(II) bromide (M_{FeBr2} = 215.65 g.mol^{-1}).

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

Iron(II) bromide: FeBr_{2}

n_{FeBr2} = *M*_{FeBr2} x V_{FeBr2} and n_{FeBr2} = $\frac{{\mathrm{m}}_{\mathrm{FeBr}2}}{{\mathrm{M}}_{\mathrm{FeBr}2}}$

⇒ *M*_{FeBr2} x V_{FeBr2 }= $\frac{{\mathrm{m}}_{\mathrm{FeBr}2}}{{\mathrm{M}}_{\mathrm{FeBr}2}}$

⇒ V_{FeBr2 }(in L) = $\frac{{\mathrm{m}}_{\mathrm{FeBr}2}(\mathrm{in}\mathrm{g})}{{\mathrm{M}}_{\mathrm{FeBr}2}(\mathrm{in}\mathrm{g}.{\mathrm{mol}}^{-1})}$ x *M*_{FeBr2} (in mol.L^{-1})

⇒ V_{FeBr2 }(in L) = $\frac{4.23}{215.65\times 0.145}$ = 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 H_{2}SO_{4}?

Equation: 2 KOH (aq) + H_{2}SO_{4} (aq) → K_{2}SO_{4} (aq) + 2 H_{2}O (l)

2 mole of KOH is required to neutralized 1 mole of H_{2}SO_{4}

⇒ when the endpoint of the neutralization is reached: n_{KOH} = 2 x n_{H2SO4}

⇒ *M*_{KOH} x V_{KOH} = 2 x *M*_{H2SO4} x V_{H2SO4}

⇒ V_{KOH} = $\frac{2\times {M}_{H2SO4}\times {\mathrm{V}}_{\mathrm{H}2\mathrm{SO}4}}{{M}_{KOH}}$

⇒ V_{KOH} = $\frac{2\times 0.120\times 200.0}{0.150}$ = 320 mL