Esercizi Svolti - Chimica

Calcoli Massa / Peso Molecolare

Esercizio 1

Quanti moli sono contenuti in 10g di Cu?

P.m. (Cu) = 63,55 g/mol

n = ^10g/_{63,55 g/mol} = 0,16 mol

Esercizio 2

Quanti atomi sono contenuti in 156,4 g di K?

P. mol (K) = 39,10 g/mol

n = ^156,4g/_{39,10 g/mol} = 4 mol

n. atomi = n • N.A. = 4 mol • 6,023 • 10²³ = 2,4 • 10²⁴ atomi

Esercizio 3

A quante mol corrispondono 50,8 g di H₂O?

P.m. (H₂O) = 2 (1,008 g/mol) + 16 g/mol = 18,016 g/mol

n = ^50,8g/_{18,016 g/mol} = 2,825 mol

Esercizio 4

A quante mol corrispondono 132 g di H₂SO₄?

P.m. (H₂SO₄) = 2 (1,008 g/mol) + 32,07 g/mol + 4 (16 g/mol) = 98,086 g/mol

n = ^132g/_{98,086 g/mol} = 1,3454 mol

Esercizio 5

Calcola il n. di molecole contenute in 10^-6 moli di CH₃CH₂OH

N. molecole = 10^-6 • 6,022 • 10²³ = 6,022 • 10¹⁷ molecole

Esercizi

1. m = ?

m = 0,2 mol di Mg(OH)₂

pmol (Mg(OH)₂) = 24,31 g/mol + 2(16 g/mol) + 2(1,008 g/mol) = 58,326 g/mol

m = n pmol

m = n pmol m = 0,2 mol · 58,326 g/mol = 11,66 g

2. g = ?

2,5 mol di H₂O

pmol (H₂O) = 2,00 g/mol + 2·35,453 g/mol + 46 g/mol = 52,458 g/mol

m = n pmol

m = 2,5 mol · 52,458 g/mol = 131,145 g

3. g = ?

0,047 mol di Al₂O₃

pmol (Al₂O₃) = 26,981 g/mol + 3(35,458 g/mol) = 133,33 g/mol

m = n pmol

m = 0,047 mol · 133,33 g/mol = 6,26 g

4. 0,000125 mol; p mol = ?

m = m

pmol = m

pmol = 0,0005 p p = 600 g/mol

0,000125 mol

5. n molcal = ?

m = 3,011·10²³ molcal di N₂

m = m molcal

m = 3,011·10²³ = 9,96·10^-4

m = NA = 3,022·10²³

6. n molcal = ?

n = 3,5·10^-1 (CH₄)

n molcal = m NA

n molcal = 3,5·10^-1 · 6,022·10²³ = 2,107·10²³

7. n atomi = ?

m = 2 g (Ar)

m = m

pmol = 137 g/mol

m = 2 g = 0,010

194 g/mol

n atomi = m NA

n atomi = 0,010 · 6,022·10²³ = 6,022·10²¹

Esercizio 4

P_mol = 162,23 g/mol

C = 59,23 g

H = 6,86 g

N = 13,84 g

m_C = 59,23 g / 4,93 mol

m_H = 6,86 g / 1,008 g/mol = 6,8 mol

m_N = 13,84 g / 14,01 g/mol = 0,98 mol

N_0,98 C_4,93 H_6,8 = N_C5H₇ = p minore

η = 162,23 g/mol : 2

η (NC₅H₇) — N₂C₁₀H₁₆ - p mol.

2)

• Na₂SO₄ -> 1 mol Na₂SO₄
• 2 mol Na p mol: = 22,98 g/mol
• 1 mol S p mol = 32 g/mol
• 4 mol O p mol = 16 g/mol

M(z) = 2 mol x 22,989 g/mol = 45,986 g

M(s) = 1 mol x 32 g/mol = 32 g

M(o) = 4 mol x 16 g/mol = 64 g

% Na = ^{45,986 g}/_{142 g} x 100 = 32%

% S = ^{32 g}/_{142 g} x 100 = 22%

% O = ^{64 g}/_{142 g} x 100

4)

• Cr₂(SO₄)₂
• 2 mol Cr p mol = 52 g/mol
• 2 mol S p mol = 32 g/mol
• 8 mol O p mol = 16 g/mol

m = 104 g

m = 64 g

m = 128 g

= 296,16 g

% Cr = ^{104 g}/_{296,16 g} x 100 = 35,13%

% S = ^{64 g}/_{296,16 g} x 100 = 21,6%

% O = ^{64 g}/_{296,16 g}

6)

• CO₂ ->
• 1 mol C p mol = 12,01 g/mol
• 2 mol O p mol = 32 g/mol

m = 12,01 g

m = 64 g

% C = ^{12,01 g}/_{76,01 g} x 100 = 27%

% O = ^{64 g}/_{76,01 g} x 100 = 73%

8)

• K 46,87%
• S 18,33%
• O 36,73%

Bdlc = 100 g

m_x = ^{46,87 g}/_{39,10 g/mol} = 1,15 mol

m_s = ^{18,38 g}/_{32,07 g/mol} = 0,57 mol

m_o = ^{36,39 g}/_{16 g/mol} = 2,28 mol

K _1,15 S _0,57 O _2,28

K₂SO₄

C = 49% H = 2% Cl = 8.1% Bdc = 1 qq

m_c = 49 / 12.01 g/mol 4.08 mol m_h = 2 g / 1.08 g/mol = 2.87 mol

m_cl = 8.1 g / 35.5 g/mol 1.36 mol

C₃H₂Cl

6.6% H 40% C 53.4% O

pmol = 80 g/mol Bdc = 1 qq

m_h = 6.6 g / 1.08 g/mol 6.55 mol m_c = 40 g / 12.01 g/mol 3.33 mol

m_o = 53.6 g / 16 g/mol 3.33 mol

H_6.6C₃O_3.3 → H₂CO

m = 80 g/mol

m(H₂CO) = H₆C₃O₃

21.43% H 53.27% O 38.24% P pmol = 161.87 uma

Bdc = 1 qq

m_h = 2.43 g /1.08 g/mol = 2.67 m_o = 53.27 g / 16 g/mol 3.3 mol

m_p = 38.24 g / 31.97 g/mol 1.23 mol

H_2.4O_3.7P_1.23 → H₂O₃P m = 161.87 uma = 2 80.946 uma

m(H₂O₃P) = H₄O₆P₂

_{red (Se²⁻)}

Se ⁰ ₊₂ →

MnO₄⁻ + Se²⁻ + H⁺ → Mn²⁺ + Se → SOm + H₂O

x -8 = -1

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O x2

5Se²⁻ → 5Se⁰ + 10e⁻

2MnO₄⁻ + 16H⁺ + 10e⁻ → 2Mn²⁺ + 8H₂O

2MnO₄⁻ + 16H⁺ + 10e⁻ + 5Se²⁻ → 2Mn²⁺ + 8H₂O + 5Se⁰ + 10e⁻

Na + H₂O₂ → Na⁺ + OH⁻ + H₂

Na → Na⁺ + 1e⁻ x2

2H₂O + 2e⁻ → H₂ + 2OH⁻

2Na + 2H₂O → 2Na⁺ + H₂ + 2OH⁻

Br₂O₃ →

red (Br⁻) Br, BrO₃⁻ ^⁵/₂ → OH⁻

oxi (Br⁻)

Br₂O₃ + H₂O →

Br₂O₃ + 2H⁺ + 2e⁻ ; F⁻ + H₂ =

2H₂O + F₂ → 2F⁻

2F⁻

Br₂O₃ + H₂O + F₂ → Br₂O₃₄ + H₂O + 2F⁻

densità H₂O = 1000 kg/m³ = 1 kg/l

1 m³ = 1000 l

1 dm³ = 1 l

1 cc = 1 ml

- Prontità in massa del glucosio

%m/m gluc: m_gluc/m_gluc+m_H2O

- 40% V/V Qual è il titolo in mol e molal in un bexch. di 250 m

%V/V V_tot = 500 ml

V_S = (40/100) ⋅ (250 ml)

Calol le mol: del glucosio con x_gluc = 0,15

x_j = n_j/n_T

mol = 0,15/(0,85)(18) ⋅ 10^-3

Calol le molal in un solor 0,273 molal H₂O in H₂O

m = 0,273 kg

Calol le mos del solero in 1,011₃ g/l

m₂ = 0,273 kg

d_soluz= 1,011₃ g/l

V_soluz = (m_soluz/d)

V_soluz = 1 mol

y = 0,271 91

I₂ + H₂SO₄ → I₂ + SO₂

2I⁻ → I₂ + 2e⁻ x 2

H₂SO₄ + 4H⁺ + 2e⁻ → SO₂ + 2H₂O x 1

5I⁻ + H₂SO₄ + 4H⁺ + 2e⁻ → I₂ + 2e⁻ + SO₂ + 2H₂O

Cr → Cr₂O₇^2- + Bi₃

2Cr³⁺ + 7H₂O → Cr₂O₇^2- + 14H⁺ + 6e⁻

Bi₂O₃ + 6H⁺ + 2e⁻ → 2Bi³⁺ + 3H₂O x 3

2Cr³⁺ + 3Bi₂O₃ + 6H⁺ → Cr₂O₇^2- + 3Bi³⁺ + 2H₂O

m(Na₂Cr₂O₇) = 0.1145

m = 0.2614 mol

5ClO₋ : 2IO₃^2-