EXAMEN.

UNIVERSIDAD NACIONAL AUTÓNOMA DE MÉXICO

FACULTAD DE ESTUDIOS SUPERIORES “ZARAGOZA”

EXAMEN ORDINARIO “A” DE LABORATORIO DE CIENCIA BASICA UNIDAD No III

NOMBRE DEL ALUMNO: Pérez García Emanuel

GRUPO: 2103 PROFESORA: Consuelo Bautista Aragón

1.-Balancear por el método ión electrón la siguiente reacción.

Br2 + (CrSCN)2+ (Br) - + (NO3) - + (SO4)2 - + CO2 + (Cr2O7 )2-

Br2 +2ē 2(Br) – Se oxida

2Cr3+ – 6ē 2Cr6+ Se reduce

3( Br2 +2ē 2(Br) – )

1( 2Cr3+ – 6ē 2Cr6+ )

3Br2 + 2(CrSCN)2+ 3(Br) - + (NO3) - + (SO4)2 - + CO2 + (Cr2O7 )2-

3Br2 + 2(CrSCN)2+ 3(Br) - + 2(NO3) - + 2(SO4)2 - + 2CO2 + (Cr2O7 )2-

Se cuentan los oxigenos para agregar H2O de lado del que falten

3Br2 + 2(CrSCN)2+ 3(Br) - + 2(NO3) - + 2(SO4)2 - + 2CO2 + (Cr2O7 )2

6O + 8O + 4O + 7O = 25O.

2.-Si se hace reaccionar diferentes volúmenes de una solución 0.025M de nitrato de estroncio con diferentes volúmenes de una solución de Sulfato de Sodio 0.025M

Tubo mL Sr(NO3)2 mL Na2SO4

1 9 1

2 6 4

3 5 5

4 4 6

5 1 9

Indique en que tubo se obtiene la mayor cantidad de sulfato e estroncio y en cada reacción la cantidad del reactivo limitante

3.-Resuelva correctamente los siguientes reactivos e incluya el procedimiento y los cálculos (S) empleado (S) para llegar a la solución.

Sr(NO3)2 + Na2SO4 SrSO4 + 2NaNO3

211.63g/mol 142g/mol 183.68g/mol

Formula: m soluto= (mol/L)(L)(g/mol)

Tubo 1

• m soluto= (0.025mol/L)(0.009L)(211.6g/mol)= 0.0476g

211.6g183.6g 183.6g 1mol

0.0476g 0.0413g 0.0413g0.00022mol

• m soluto= (0.025mol/L)(0.001L)(142g/mol)= 0.0035g

142g183.6g 142g 1mol

0.0035g 0.0045g 0.0045g0.00003mol

Tubo 2

• m soluto= (0.025mol/L)(0.006L)(211.6g/mol)= 0.0317g

211.6g183.6g 183.6g 1mol

0.0317g 0.0275g 0.0275g0.00014mol

• m soluto= (0.025mol/L)(0.004L)(142g/mol)= 0.0142g

142g183.6g 142g 1mol

0.0142g 0.0183g 0.0183g0.00012mol

Tubo 3

• m soluto= (0.025mol/L)(0.005L)(211.6g/mol)= 0.0264g

211.6g183.6g 183.6g 1mol

0.0264g 0.0229g 0.0229g0.00012mol

• m soluto= (0.025mol/L)(0.005L)(142g/mol)= 0.0177g

142g183.6g 142g 1mol

0.0177g 0.0228g 0.0228g0.00016mol

Tubo 4

• m soluto= (0.025mol/L)(0.004L)(211.6g/mol)= 0.0211g

211.6g183.6g 183.6g 1mol

0.0211g 0.0183g 0.0183g0.00009mol

• m soluto=

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