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malfutka [58]
3 years ago
5

Teniendo en cuenta la siguiente reacción quimica: Mn (NO3)2 + NaBiO; + HNO, HMnO₂+ Bi (NO3)3 + NaNO3 + H₂O ¿Cuánto Bi (NO₂), se

obtiene a partir de 650 g de Mn (NO₂)2 y 700 g de NaBiO, si la eficiencia del proceso es de un 80%?
Chemistry
1 answer:
o-na [289]3 years ago
7 0

Answer:

La masa de Bi (NO₃) ₃ producida es de aproximadamente 2.117,09 gramos

Explanation:

La ecuación química de la reacción se presenta como sigue;

2Mn (NO₃) ₂ + 5NaBiO₃ + 14HNO₃ → 2HMnO₄ + 5Bi (NO₃) ₃ + 5NaNO₃ + 7H₂O

La masa dada de Mn (NO₃) ₂ = 650 g

La masa de NaBiO₃ = 750 g

El número de moles de Mn (NO₃) ₂ = 650 g / (178.95 g / mol) ≈ 3.63 moles

El número de moles de NaBiO₃ = 750 g / (279.968 g / mol) ≈ 2.68 moles

Por lo tanto, el NaBiO₃, es el reactivo limitante, y el rendimiento teórico de Bi (NO₃) ₃ ≈ 5/2 × 2.68 = 6.7 moles

Dado que la eficiencia del proceso es del 80%, el rendimiento real de Bi (NO₃) ₃ = 0,8 × 6,7 moles = 5,36 moles de Bi (NO₃) ₃

La masa molar de Bi (NO₃) ₃ ≈ 394,98 g / mol

La masa de Bi (NO₃) ₃ = 394,98 g / mol × 5,36 moles ≈ 2,117,09 g

La masa de Bi (NO₃) ₃ produjo ≈ 2,117.09 g

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