Necesito ayuda urgente con mi tarea de química porfavor
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1.
C: 52.2 g/ 12 g/mol = 4.35 mol
O: 34.8 g / 16 g/mol = 2.175 mol
H: 13 g/ 1 g/mol = 13 mol
dividir entre el menor de los resultados cada elemento
C: 4.35 mol/ 2.175 mol = 2
O: 2.175 mol/2.175 mol = 1
H:13 mol / 2.175 mol = 6
FE: C2H6O
Mm FE
C: 2 x 12 = 24 g/mol
H: 6 x1 = 6 g/mol
O: 1 x 16 = 16 g/mol
````````````````````````````
Mm = 46 g/mol
n= 92 g/mol / 46 g/mol = 2
FM = (C2OH6)2 = C4O2H12
2.
Ba: 52.5 g/ 137 g/mol = 0.383 mol
N: 10.7 g/ 14g/mol = 0.764 mol
O: 36.8 g/ 16 g/mol = 2.3 mol
dividir
Ba: 0.383 mol / 0.383 mol = 1
N: 0.764 mol / 0.383 mol = 2
O: 2.3 mol / 0.383 mol = 6
FE: BaN2O6
MmFE
Ba: 1 x 137 = 137 g/mol
N: 2 x 14 = 28 g/mol
O: 6 x 16 = 96 g/mol
``````````````````````````````
Mm = 261 g/mol
n = 261 g/mol/261 g/mol = 1
FM= (BaN2O6)1 = BaN2O6
3.
1.CALCULAR MOLES de Ca(OH)2 Y DE H3PO4
n Ca(OH)2 = 35 g/ 74 g/mol = 0.473 mol
n H3PO4 = 57 g / 98 g/mol = 0.582 mol
2. calcular mol de Ca3(PO4)2 a partir de los moles calculados
mol Ca3(PO4)2 = 0.473 mol Ca(OH)2 X 1 mol Ca3(PO4)2
``````````````````````````
3 mol Ca(OH)2
mol de Ca3PO4)2 = 0.158 mol
mol Ca3(PO4)2 = 0.582 mol H3PO4 X 1 mol Ca3(PO4)2
``````````````````````````
2 mol H3PO4
mol de Ca3PO4)2 = 0.291 mol
RL: Ca(OH)2. es aquel que origina la menor cantidad de moles de producto (0.158 moles de Ca3(PO4)2)
4.
g MgCl2 = 30 g Mg x 1mol Mg x 1 mol MgCl2 x 95 g MgCl2
```````````` ```````````````` ``````````````````
24 g Mg 1 mol Mg 1 mol MgCl2
g = 118.75 g de MgCl2
C: 52.2 g/ 12 g/mol = 4.35 mol
O: 34.8 g / 16 g/mol = 2.175 mol
H: 13 g/ 1 g/mol = 13 mol
dividir entre el menor de los resultados cada elemento
C: 4.35 mol/ 2.175 mol = 2
O: 2.175 mol/2.175 mol = 1
H:13 mol / 2.175 mol = 6
FE: C2H6O
Mm FE
C: 2 x 12 = 24 g/mol
H: 6 x1 = 6 g/mol
O: 1 x 16 = 16 g/mol
````````````````````````````
Mm = 46 g/mol
n= 92 g/mol / 46 g/mol = 2
FM = (C2OH6)2 = C4O2H12
2.
Ba: 52.5 g/ 137 g/mol = 0.383 mol
N: 10.7 g/ 14g/mol = 0.764 mol
O: 36.8 g/ 16 g/mol = 2.3 mol
dividir
Ba: 0.383 mol / 0.383 mol = 1
N: 0.764 mol / 0.383 mol = 2
O: 2.3 mol / 0.383 mol = 6
FE: BaN2O6
MmFE
Ba: 1 x 137 = 137 g/mol
N: 2 x 14 = 28 g/mol
O: 6 x 16 = 96 g/mol
``````````````````````````````
Mm = 261 g/mol
n = 261 g/mol/261 g/mol = 1
FM= (BaN2O6)1 = BaN2O6
3.
1.CALCULAR MOLES de Ca(OH)2 Y DE H3PO4
n Ca(OH)2 = 35 g/ 74 g/mol = 0.473 mol
n H3PO4 = 57 g / 98 g/mol = 0.582 mol
2. calcular mol de Ca3(PO4)2 a partir de los moles calculados
mol Ca3(PO4)2 = 0.473 mol Ca(OH)2 X 1 mol Ca3(PO4)2
``````````````````````````
3 mol Ca(OH)2
mol de Ca3PO4)2 = 0.158 mol
mol Ca3(PO4)2 = 0.582 mol H3PO4 X 1 mol Ca3(PO4)2
``````````````````````````
2 mol H3PO4
mol de Ca3PO4)2 = 0.291 mol
RL: Ca(OH)2. es aquel que origina la menor cantidad de moles de producto (0.158 moles de Ca3(PO4)2)
4.
g MgCl2 = 30 g Mg x 1mol Mg x 1 mol MgCl2 x 95 g MgCl2
```````````` ```````````````` ``````````````````
24 g Mg 1 mol Mg 1 mol MgCl2
g = 118.75 g de MgCl2
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