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Andrews [41]
3 years ago
7

Calcule el volumen de disolución de LiOH a 3.5 M, necesario para neutralizar una disolución de 25 ml de H2SO3 cuya densidad es d

e 1.03 g/ml y su riqueza del 60%. (60 PTS.) Alguien puede ayudarme con esto porfabor
Chemistry
1 answer:
Olin [163]3 years ago
7 0

Respuesta:

0.11 L

Explicación:

Paso 1: Escribir la ecuación balanceada

2 LiOH + H₂SO₃ ⇒ Li₂SO₃ + H₂O

Paso 2: Calcular la masa de solución de H₂SO₃

25 mL de solución de H₂SO₃ tiene una denisdad de 1.03 g/mL.

25 mL × 1.03 g/mL = 28 g

Paso 3: Calcular la masa de H₂SO₃ en 26 g de Solución de H₂SO₃

La riqueza de H₂SO₃ es 60%, es decir, cada 100 g de solución hay 60 g de H₂SO₃.

26 g Sol × 60 g H₂SO₃/100 g Sol = 16 g H₂SO₃

Paso 4: Calcular los moles correspondientes a 16 g de H₂SO₃

La masa molar de H₂SO₃ es 82.07 g/mol.

16 g × 1 mol/82.07 g = 0.19 mol

Paso 5: Calcular los moles de LiOH que reaccionan con 0.19 moles de H₂SO₃

La relación molar de LiOH a H₂SO₃ es 2:1. Los moles de LiOH que reaccionan son 2/1 × 0.19 mol = 0.38 mol.

Paso 6: Calcular el volumen de solución de LiOH

0.38 moles de LiOH están en una solución 3.5 M. El volumen requerido es:

0.38 mol × 1 L/3.5 mol = 0.11 L

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