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
Answer: E
How much NH₃ can be produced from the reaction below:
N₂ + 3H₂ - 2NH₃
The stoichiometric ratio of the reactants = 1:3
Given
74.2g of N₂, and Molar mass = 14g/mole
Mole of N₂ = 74.2/14=5.3mols of N₂,
and 14mols of H₂
From this given values and comparing with the stoichiometric ratio, H₂ will be the limiting reagent while N₂ is the excess reactant.
i.e, for every 14mols of H₂, we need 4.67mols of N₂ to react with it to produce 9.33mols of NH₃ as shown (vice versa)
From this we have 9.33mols of NH₃ produced
Avogadro constant, we have n = no of particles = 6.022x10²³ molecules contained in every mole of an element.
For a 9.33mols of NH3, we have 9.33x6.022x10²³molecules in NH3
5.62x10²⁴molecules of NH₃
<span>According to naming rules, the types of compound that use prefixes in their names are the covalent compounds. These are formed by sharing of valence electrons, by the atoms, and results in the formation of a covalent bond. For example, water and DNA.</span>