Calculate with reaction equation 55 (st-76) Different exercises round to This Chemical Calculation. Lightweight exercises a) The Formula for Methyl alcohol is CH3OH. What ? Molecular mass owned a such as molecule? b) How many Oxygen atoms contained 1 g O2-Moleküle? c) How many Atoms contains 1g Helium gas? d) How a lot Grams corresponding to 1 mol Sulphur, 1 mol Sodium and 1 mol Water? e) How Large is The Weight of: 0.5 mol Carbon dioxide (Formula CO2) 0.1mol Sulphuric acid (Formula H2SO4)? f) How Large is The Molecular mass One of them Substance, where: 0.87 mol of you 24.36 g Weighing? g) How a lot C-Atoms contained 24 g Carbon? h) How many Oxygen atoms contained 1 g Ozone (O3)? Indication in Mol (n) and 'real' Number of Half weight exercises i) How a lot Grams Water It is The Commission In the case of The Incineration of 100 g Methane (CH4)? j) Men and boys (Pb304) It's not The Commission from Lead oxide (PbO) by: Oxidation with Oxygen represent. Calculate The Weight to Lead oxide, The to the Manufacture of 100 kg Men and boys Required is. k) Alcohol (C2H5OH) will be Usually by: Fresh or chilled of Glucose (C6H12O6) Presented, accordingly The reaction equation : Glucose molecule –> 2 Alcohol molecules + 2 Carbon dioxide molecules How a lot kg Alcohol to leave The Commission from 2 kg Glucose winning? Weight exercises l) How a lot Grams Zinc sulphide(ZnS) received We are In the case of The Implementation of 200 g Zinc with 1000 g Sulphur (S8)?

  Aufgabe: st-76    

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Different exercises round to This Chemical Calculation.

1. Lightweight exercises
2. a) The Formula for Methyl alcohol is CH₃OH. What ? Molecular mass owned a such as molecule?
3. b) How many Oxygen atoms contained 1 g O₂-Moleküle?
4. c) How many Atoms contains 1g Helium gas?
5. d) How a lot Grams corresponding to 1 mol Sulphur, 1 mol Sodium and 1 mol Water?
6. e) How Large is The Weight of:
       0.5 mol Carbon dioxide (Formula CO₂)
       0.1mol Sulphuric acid (Formula H₂SO₄)?
7. f) How Large is The Molecular mass One of them Substance, where: 0.87 mol of you 24.36 g Weighing?
8. g) How a lot C-Atoms contained 24 g Carbon?
9. h) How many Oxygen atoms contained 1 g Ozone (O₃)? Indication in Mol (n) and 'real' Number of
10. Half weight exercises
11. i) How a lot Grams Water It is The Commission In the case of The Incineration of 100 g Methane (CH₄)?
12. j) Men and boys (Pb₃0₄) It's not The Commission from Lead oxide (PbO) by: Oxidation with Oxygen represent. Calculate The Weight to Lead oxide, The to the Manufacture of 100 kg Men and boys Required is.
13. k) Alcohol (C₂H₅OH) will be Usually by: Fresh or chilled of Glucose (C₆H₁₂O₆) Presented, accordingly The reaction equation :
        Glucose molecule –> 2 Alcohol molecules + 2 Carbon dioxide molecules
        How a lot kg Alcohol to leave The Commission from 2 kg Glucose winning?
14. Weight exercises
15. l) How a lot Grams Zinc sulphide(ZnS) received We are In the case of The Implementation of 200 g Zinc with 1000 g Sulphur (S₈)?

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Lösung:

1. Lightweight exercises
2. a) m(CH₃OH) = 12 u + 3·1u + 16u + 1u = 32 u
3. b) n(O₂)= m(O₂)/M(O₂) = 1 g / 32 g/mol = 1/32 mol
       n(O) = 1/16 mol (= 3.76·10²²)
4. c) n(He) = m(He)/M(He) = 1g/4g/mol = 0.25 mol (= 1.5·10²³)
5. d) 1 mol S = 32 g, 1 mol Na = 23 g, 1 mol H₂O = 18 g
6. e) 1 mol CO₂ = 44 g, 0.5 mol: 22 g
       1 mol H₂SO₄: 98 g, 0.1 mol: 9.8 g
7. f) n=m/M; M = m/n = 24.36 g/ 0.87 mol = 28 g/mol Responsibility. 28 u
8. g) n=m/M; n(C) = m(C)/M(C) = 24g/12g/mol = 2 mol (=1.2·10²⁴)
9. h) n(O₃)=m(O₃)/M(O₃) = 1/48 mol
       n(O) = 3/48 mol = 1/16 mol (=3.76·10²²)
10. Half weight exercises
11. i) CH₄ + 2·O₂ ⇄ CO₂ + 2·H₂O
        Look at this. of which: Table !
        Es will be So what do you mean? 18·12.5 = 225 Grams Water of which:.
12. j) 6·PbO + O₂ ⇄ 2·Pb₃0₄
        Look at this. of which: Table !
        Es will be So what do you mean? about. 97.6 kg PbO Required
13. k) C₆H₁₂O₆ ⇄ 2·C₂H₅OH + 2·CO₂
        Look at this. of which: Table !
        Es will be about. 1.02 kg Ethanol received
14. Weight exercises
15. l) 8·Zn + S₈ ⇄ 8 ZnS
        · The exercise is over-determined, two Situations Calculation
        · Look at this. of which: Table !
        · l-1: It 's just … with (200) g Zinc
        · l-2: It 's just … with (1000g) Sulphur
        · Please note:, that with 200 g Zinc and At the same time 1000 g S Started will be
        · Variation l-2 would be 2043 g Zinc needing, of which: are But It 's just … 200 g
        · Variation l-1 would be All of it Zinc use up, However, remaining S₈ remaining

Manufacture	M(g/mol)	m(g)	n(mol)
i) C4H	16	100	100/16=6.25
i) CO2	18	225	2·6.25 = 12.5
j) Pb3O4	685.6	100'000	100'000/685.6=145.9
j) PbO	223.2	97'605	3·145.9=437.6
k) C6H12O6	180	2000	11.11
k) C2H5OH	46	1022	2·11.11 = 22.22
l-1) Zn	65.4	200	3.05
l-1) S8	256	97.6	3.05/8=0.38
l-1) ZnS	97.4	297.04	3.05
l-2) S8	256	1000	3.9
l-2) Zn	65.4	2043	8·3.9=31.25
l-2) ZnS	97.4	3043	31.25

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