PLAN DE MEJORAMIENTO QUIMICA
Enviado por kilyam sebastian acosta pineda • 22 de Noviembre de 2021 • Exámen • 824 Palabras (4 Páginas) • 54 Visitas
PLAN DE MEJORAMIENTO QUIMICA
PUNTO 1
- 6KOH + 3 Cl2 → KClO3 + 5KCl + 3H2O
6 mol 3 mol 1 mol 5 mol 3 mol
10 mol KOH * 1 mol KClO3 / 6 mol KOH = 1.67 mol KClO3
5 mol KOH * 3 mol Cl2 / 6 mol KOH= 2.5 mol Cl2
- A partir de 6 moles de KOH se produce 1 mol de KClO₃
A partir de 10 moles de KOH.................. X
X = ( 10 mol KOH × 1 mol KClO₃) / 6 mol KOH = 1,7 moles KClO₃
- 6 moles de KOH reaccionan con 3 moles de Cl₂
5 moles de KOH ......................... X
X = ( 5 mol KOH × 3 mol Cl₂) / 6 mol KOH = 2,5 moles de Cl₂
PUNTO 2
S + Cu → CuS
Mm S = 32 g/mol
Cu = 63.54 g/mol
CuS = 95.54 g/mol
1. calcular moles de S
mol CuS = 64 g S x 1 mol S x 1 mol CuS
```````````` ``````````````
32 g S 1 mol CuS
mol CuS = 2
2. calcular gramos de CuS
g CuS = 64 g S x 1 mol S x 1 mol CuS x 95.54 g CuS
```````````` ``````````````` ``````````````````
32 g S 1 mol S 1 mol CuS
g CuS = 191.08 g
PUNTO 3
- 2 H2 + O2 → 2 H2O
- 1 mol de H₂O
100 gr de H · ------------------------ = 50 moles de H₂O
2 gr de hidrógeno
18 gr H₂O
50 moles H₂O · -------------------------- = 900 gr de H₂O
1 mol H₂O
PUNTO 4
- Bi2S3+9/2º2 ---- Bi2o3+3SO2
- [pic 1]
[pic 2]
[pic 3]
- [pic 4]
PUNTO 5
- [pic 5]
- g H2 = 68 g NH3 X 1 mol NH3 X 3 mol H2 x 2 g H2
``````````````` ````````````` ````````````
17 g NH3 2 mol NH3 1 mol H2
g = 1.2 de H2
- mol NH3 = 68 g/ 17 g/mol = 4 mol
H2 = 4 mol NH3 x 3 mol H2 x 6.022 x 10²³ moléculas
2 mol NH3 1 mol H2
= 3.61 x 10²⁴
...