Answer :
Final answer:
To find the mass of lead in 139 g of Pb(NO3)2, we calculate the molar mass of Pb(NO3)2, find the number of moles of Pb(NO3)2, and then use the molar mass of lead to find its mass in the compound, resulting in approximately 87.02 g of lead.
Explanation:
The question asks to calculate the mass of lead in 139 g of Pb(NO3)2. First, we need to calculate the molar mass of Pb(NO3)2. The molar mass of lead (Pb) is 207.2 g/mol, and for nitrate (NO3) - since nitrogen (N) has a molar mass of 14.0 g/mol and oxygen (O) has 16 g/mol - a single nitrate ion would have a molar mass of 62 g/mol (1×14.0 + 3×16.0). As there are two nitrate ions per molecule of lead nitrate, the total molar mass of NO3 in Pb(NO3)2 is 124 g/mol. Thus, the total molar mass of Pb(NO3)2 is 207.2 g/mol (for Pb) + 124 g/mol (for 2×NO3), totaling 331.2 g/mol.
To find the mass of lead in the given 139 g of Pb(NO3)2, we follow these steps:
Calculate the number of moles of Pb(NO3)2 in 139 g:
number of moles = mass (g) / molar mass (g/mol)
= 139 g / 331.2 g/mol
≈ 0.42 moles.
Since the ratio of lead to Pb(NO3)2 in terms of moles is 1:1, there are also approximately 0.42 moles of lead in 139 g of Pb(NO3)2.
Calculate the mass of lead: mass = number of moles × molar mass of lead
= 0.42 moles × 207.2 g/mol
= 87.02 g.
Therefore, the mass of lead in 139 g of Pb(NO3)2 is approximately 87.02 g.
