Calculate with reaction equation 7 (st-78) Calculate The Masses of all Participants Particles the following: reaction equations PbO + C → Pb + CO2, given be 0.528 g C. M(PbO) = 223.2 g/mol, M(Pb) = 207.2 g/mol, M(CO2) = 44 g/mol NaCl → Na + Cl2, given be 100 g Na. M(NaCl) = 58.5 g/mol, M(Cl2) = 71 g/mol Fe + O2 → Fe2O3, given be 100 Of which: Iron and steel. M(O2) = 32 g/mol, M(Fe2O3) = 159.6 g/mol Sodium (Na) and Chlorinated gases (Cl2) to be created by: One Fusion electrolysis (special Technical Procedures) from Sodium chloride (NaCl). Place The reaction equation on and Calculate, How a lot Grams Sodium chloride used will be must be, to 100.0 g Sodium to received. Calcium carbide (CaC2) is One Solids, The with Water a Containing by weight: Gas developed, has been Previously also for Machinery for the manufacture of lamps Required. reaction equation: Ca + 2·C → CaC2 How a lot kg CaC2 receives It's, where: It's 50 kg Ca used? How a lot kg Carbon will be For that Required.

  Aufgabe: st-78    

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Calculate The Masses of all Participants Particles the following: reaction equations

1. PbO + C → Pb + CO₂, given be 0.528 g C.
       M(PbO) = 223.2 g/mol, M(Pb) = 207.2 g/mol, M(CO₂) = 44 g/mol
2. NaCl → Na + Cl₂, given be 100 g Na.
       M(NaCl) = 58.5 g/mol, M(Cl₂) = 71 g/mol
3. Fe + O₂ → Fe₂O₃, given be 100 Of which: Iron and steel.
       M(O₂) = 32 g/mol, M(Fe₂O₃) = 159.6 g/mol
4. Sodium (Na) and Chlorinated gases (Cl₂) to be created by: One Fusion electrolysis (special Technical Procedures) from Sodium chloride (NaCl). Place The reaction equation on and Calculate, How a lot Grams Sodium chloride used will be must be, to 100.0 g Sodium to received.
5. Calcium carbide (CaC₂) is One Solids, The with Water a Containing by weight: Gas developed, has been Previously also for Machinery for the manufacture of lamps Required.
       reaction equation: Ca + 2·C → CaC₂
       How a lot kg CaC₂ receives It's, where: It's 50 kg Ca used? How a lot kg Carbon will be For that Required.

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Lösung:

1. · Es Apply the following: Formulas:
       n=m/M Responsibility. m = n·M
2. · Look at this. of which: Table
3. a) Please note:, that The Equation Not yet Balanced will be must:
       2·PbO + C → 2·Pb + CO₂
       The Searched Masses amount:
       · m(PbO) = 19.64 g
       · m(Pb) = 18.23 g
       · m(CO₂) = 1.94 g
4. b) Please note:, that The Equation Not yet Balanced will be must:
       2·NaCl → 2·Na + Cl₂
       The Searched Masses amount:
       · m(NaCl) = 254.4 g
       · m(Cl₂) = 154.4 g
5. c) Please note:, that The Equation Not yet Balanced will be must:
       4·Fe + 3·O₂ → 2·Fe₂O₃
       Please note:, that 100 Of which: Iron and steel given are. The Accounts is in g/mol.
       m(O₂) = about. 42.9 Of which:
       m(Fe₂O₃) = about. 142.7 Of which:
6. d) Identical to exercise b). reaction equation:
       2·NaCl → 2·Na + Cl₂
7. e) The Searched Masses amount:
       · Please note:, that The Information to be provided in kg are. Conversion on g
       · m(CaC₂) = 80 kg
       · m(C) = 30 kg

Manufacture	M(g/mol)	m(g)	n(mol)
a) PbO	223.2	19.64	2·0.044=0.088
a) C	12	0.528	0.528/12 = 0.044
a) Pb	207.2	18.23	2·0.044 = 0.088
a) CO2	44	1.94	1·0.044 = 0.044
–––	–––	–––	–––
b) NaCl	58.5	254.4	4.35
b) Na	23	100	100/23 = 4.35
b) Cl2	71	154.4	4.35/2 = 2.174
–––	–––	–––	–––
c) Fe	55.9	108	108/55.9 = 1'788'908
c) O2	32	42'933'810.4	1'788'908/4·3= 1'341'681
c) Fe2O3	159.6	142'754'919.5	1'788'908/2= 894'454
–––	–––	–––	–––
d) NaCl	58.5	254.4	4.35
d) Na	23	100	100/23 = 4.35
d) Cl2	71	154.4	4.35/2 = 2.174
–––	–––	–––	–––
e) Ca	40	50'000	50'000/4=1250
e) C	12	30'000	2·1250=2500
e) CaC2	64	80'000	1·1250=1250

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