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
Answer:
I think it's A,C,E.
Explanation:
I'm not 100% sure, but I think their right, hopefully.
Answer:
C) 2H2SO4(aq)+ 2Ba(OH)2(aq)---->2BaSO4(s)+4H2O(I)
Explanation:
The reaction proper:
2H₂SO₄ + 2Ba(OH)₂ → 2BaSO₄ + 4H₂O
The above reactions is a typical acid-base neutralization reaction.
A reaction in which oxidation-reduction did not take place is called a non-redox reaction. In such a reaction, there is no transfer of electrons between reactants, that is, none of the atoms of the reactants changes its oxidation number in forming the products.
Most neutralization reactions are non-redox reactions.