Assuming that the reaction is quantitative, calculate the mass of sulfur (S)obtained by reacting 3.0 g of PbS with 2.0 g of HNO 3 (hint: identify thelimiting reagent first).

To solve this problem, we first need to write down the balanced chemical equation for the reaction. The reaction of lead sulfide (PbS) with nitric acid (HNO3) produces lead nitrate (Pb(NO3)2), sulfur (S), and water (H2O). The balanced equation is:

PbS + 2HNO3 → Pb(NO3)2 + S + H2O

Next, we need to calculate the number of moles for PbS and HNO3. The molar mass of PbS is approximately 239.3 g/mol and the molar mass of HNO3 is approximately 63.0 g/mol.

Number of moles of PbS = mass/molar mass = 3.0 g / 239.3 g/mol = 0.0125 mol Number of moles of HNO3 = mass/molar mass = 2.0 g / 63.0 g/mol = 0.0317 mol

From the balanced equation, we can see that the mole ratio of PbS to HNO3 is 1:2. This means we need 2 moles of HNO3 for every mole of PbS. However, we have less than 2 moles of HNO3 for every mole of PbS (0.0317 mol < 2*0.0125 mol), so HNO3 is the limiting reagent.

Finally, we can calculate the mass of sulfur produced. From the balanced equation, we can see that 1 mole of HNO3 produces 1 mole of sulfur. Therefore, the number of moles of sulfur produced is the same as the number of moles of the limiting reagent, HNO3.

Number of moles of S = 0.0317 mol

The molar mass of sulfur is approximately 32.1 g/mol, so the mass of sulfur produced is:

Mass of S = number of moles * molar mass = 0.0317 mol * 32.1 g/mol = 1.02 g

So, assuming the reaction is quantitative, 1.02 g of sulfur is obtained by reacting 3.0 g of PbS with 2.0 g of HNO3.