The oxides of nitrogen are very important ingredients in determining urban air pollution. Name each of the following compounds.(a) N2O(b) NO(c) NO2(d) N2O5(e) N2O4

(a) N₂O - Dinitrogen monoxide or Dinitrogen (I) oxide

(b) NO - Nitrogen monoxide or Nitrogen (II) oxide

(c) NO₂ - Nitrogen dioxide or Nitrogen (IV) oxide

(d) N₂O₅ - Dinitrogen pentoxide

(e) N₂O₄ - Dinitrogen tetraoxide.

Explanation:

To name each of the compounds,

(a) N2O; It is properly written as N₂O. N₂O is a colorless, sweet tasting gas also known as "laughing gas".

N₂O - Dinitrogen monoxide OR Dinitrogen (I) oxide.

(b) NO; NO is a colorless gas.

NO - Nitrogen monoxide OR Nitrogen (II) oxide.

(c) NO2; It is properly written as NO₂

NO₂ is a reddish-brown gas.

NO₂ - Nitrogen dioxide or Nitrogen (IV) oxide.

(d) N2O5; It is properly written as N₂O₅

N₂O₅ is a white solid.

N₂O₅ - Dinitrogen pentoxide.

(e) N2O4; It is properly written as N₂O₄;

N₂O₄ is a red-brown liquid with an unpleasant smell.

N₂O₄ - Dinitrogen tetraoxide.

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g