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 = m sucrose M 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 = 12 100 x m solution Mass of this solution needed: m solution = 100 12 x m sucrose m solution = 100 12 x n sucrose x M sucrose m solution = 100 12 x 35.1 x 10 -3 x 342 m solution = 1.00 x 10 2 g

Midterm 1

General Chemistry 2

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Question 1

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?

Solution

Potassium hydrogen phosphate: K₂HPO₄ 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: 0.150M potassium hydrogen phosphate During a dilution: conservation of the number of moles ⇒ n₁ = n₂ ⇒ M₁x V₁ = M₂ x V₂ ⇒ V₁ =

\frac{M_{2} \times V_{2}}{M_{1}}

\frac{0.150 \times 650}{0.800}

= 1.22 x 10² mL

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

Question 2

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.

Solution

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)

Question 3

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?

Solution

n_sucrose = $\frac{m_{sucrose}}{M_{sucrose}}$ 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 10² g

Question 4

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?

Solution

1) n = $\frac{N}{N_{A}}$ and n = $\frac{m}{M}$ 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 = $\frac{m}{M}$ x N_A

⇒ 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) n_sucrose = $\frac{m_{sucrose}}{M_{sucrose}}$ = $\frac{10.0}{342}$ = 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

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

n_O2 = $\frac{m_{O 2}}{M_{O 2}}$ = $\frac{12.0}{32}$ = 3.75 x 10^-1 mol > n_{O2 needed}

The sucrose is the limiting reagent of this reaction

Question 5

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?

Solution

Iron(II) bromide: FeBr₂

n_FeBr2 = M_FeBr2 x V_FeBr2 and n_FeBr2 = $\frac{m_{FeBr 2}}{M_{FeBr 2}}$

⇒ M_FeBr2 x V_FeBr2= $\frac{m_{FeBr 2}}{M_{FeBr 2}}$

⇒ V_FeBr2(in L) = $\frac{m_{FeBr 2} (in g)}{M_{FeBr 2} (in g . {mol}^{- 1})}$ 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

Question 6

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?

Solution

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: n_KOH = 2 x n_H2SO4

⇒ M_KOH x V_KOH = 2 x M_H2SO4 x V_H2SO4

⇒ V_KOH = $\frac{2 \times M_{H 2 S O 4} \times V_{H 2 SO 4}}{M_{K O H}}$

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

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