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Alexus [3.1K]
3 years ago
15

El acido nitrico se obtiene industrialmente por oxidacion de amoniaco (nh3), de acuerdo con la ecuacion: 4nh3 + 702 - h2o + hno2

+ 2hno3 ¿ cuanto acido nitrico del 90% de pureza se podra obtener a partir de 1.36 kg de amoniaco?
Chemistry
1 answer:
devlian [24]3 years ago
6 0

Answer:

2800 g de ácido nítrico

Explanation:

La ecuación por la oxidación de amoniaco es:

4NH₃  +  7O₂  →  4H₂O  +  2HNO₂  +  2HNO₃

Si pensamos que el oxígeno es el reactivo limitante, trabajamos con el amoniaco. Convertimos su masa a moles:

1.36 kg = 1360 g

1360 g . 1mol  /17g = 80 moles

Si 4 moles de amoniaco pueden producir 2 moles de acido nítrico

80 moles producirán, (80 . 2)/4 = 40 moles.

Convertimos los moles a gramos:

40 mol . 63g /mol = 2520 g

Si le aplicamos la pureza

2520 g . 100/90 = 2800 g

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The balanced equation for the above reaction is as follows
C₆H₁₂O₆(s) + 6O₂(g) --> 6H₂O(g) + 6CO₂<span>(g)
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If a solution initially contains 0.260 M HC2H3O2, what is the equilibrium concentration of H3O+ at 25 ∘C? Express your answer in
arlik [135]

Answer:

2.16 × 10⁻³

Explanation:

Step 1: Given data

Concentration of the acid (Ca): 0.260 M

Acid dissociation constant (Ka): 1.80 × 10⁻⁵

Step 2: Write the acid dissociation equation

HC₂H₃O₂(aq) + H₂O(l) ⇄ C₂H₃O₂⁻(aq) + H₃O⁺(aq)

Step 3: Calculate the concentration of H₃O⁺ at equilibrium

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