Question

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

reacted?

Answer 1

Mass = 76.176 g

Step-by-step 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