C. If 62.9 g of lead (II) chloride is produced, how many grams of lead (II) nitrate were

reacted?

Mass = 76.176 g

Explanation:

Given dа ta:

Mass of lead (II) chloride produced = 62.9 g

Mass of lead (II) nitrate used = ?

Solution:

Chemical equation:

Pb (NO₃) ₂ + 2HCl → PbCl₂ + 2HNO₃

Number of moles of lead (II) chloride:

Number of moles = mass/molar mass

Number of moles = 62.9 g / 278.1 g/mol

Number of moles = 0.23 mol

Now we will compare the moles of lead (II) chloride with Pb (NO₃) ₂ from balance chemical equation:

PbCl₂ : Pb (NO₃) ₂

1 : 1

0.23 : 0.23

Mass of Pb (NO₃) ₂:

Mass = number of moles * molar mass

Mass = 0.23 mol * 331.2 g/mol

Mass = 76.176 g