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 data:

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

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