PLEASE HELP If 62.6 g of lead (II) chloride is produced, how many grams of lead (II) nitrate were reacted?

The mass of lead (II) nitrate required is 74.52 g

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

......(1)

Given mass of lead (II) chloride = 62.6 g

Molar mass of lead (II) chloride = 278.1 g/mol

Plugging values in equation 1:

The chemical equation for the reaction of lead (II) chloride and sodium nitrate follows:

By the stoichiometry of the reaction:

1 mole of lead (II) chloride is produced from 1 mole of lead (II) nitrate

Then, 0.225 moles of lead (II) chloride will react with = of lead(II) nitrate

Molar mass of lead (II) nitrate = 331.2 g/mol

Plugging values in equation 1:

Hence, the mass of lead (II) nitrate required is 74.52 g

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3