Answer:
If 51.8 of Pb is reacting, it will require 4.00 g of O2
If 51.8 g of PbO is formed, it will require 3.47 g of O2.
Explanation:
Equation of the reaction:
2 Pb + O2 → 2 PbO
From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO
Molar mass of Pb = 207 g
Molar mass of O2 = 32 g
Molar mass of PbO = 207 + 32 = 239 g
Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO
= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO
Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.
If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2
Five hundred twenty million, three hundred and forty thousand. hope it helps!
b) It is based on atomic properties as alkali metals requires 7 more electrons to complete their outer orbit. And they try to give those electrons to other elements to obtain noble gas configuration.
Noble gases are the gases which do not react easily with anything. They are also called as Inert gases, and belongs to group 18 of the periodic table.
Alkali metals are the substances which are found in Group I of a periodic table. Mostly the elements which are present are:
Properties of alkali metals are: Soft, shiny reactive metals. They are soft enough to cut with knife. Metals react with water and air quickly and gets tarnish, so pure metals are stored in container by dipping them in oil to prevent oxidation.
To know more about Alkali metals, refer to this link:
brainly.com/question/18153051
#SPJ4
