How many grams of lead sulfide from when 10.0 g of lead are heated with 3.0 g of sulfur

The reaction forms 11.5 g PbS.  

We have the masses of two reactants, so this is a limiting reactant problem.

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.  

Step 1. Gather all the information in one place with molar masses above the formulas and everything else below them.  

M_r:      207.2  32.06  239.28

              Pb +       S →    PbS

Mass/g: 10.0       3.0

Step 2. Calculate the moles of each reactant  

Moles of Pb  = 10.0 g Pb × (1 mol Pb /207.2 g Pb) = 0.048 26 mol Pb  

Moles of S = 30 g S × (1 mol S/32.06 g S) = 0.9357 mol S

Step 3. Identify the limiting reactant

Calculate the moles of PbS we can obtain from each reactant.  

From Pb: Moles of PbS = 0.048 26 mol Pb × (1 mol PbS/1 mol Pb )

= 0.048 26 mol PbS

From S: Moles of S = 0.9357 mol S × (1 mol PbS/1 mol S) = 0.9357 mol PbS

Pb is the limiting reactant because it gives the smaller amount of PbS.

Step 4. Calculate the mass of PbS.

Mass = 0.048 26 mol PbS × (239.28 g PbS/1 mol PbS) = 11.5 g PbS

The reaction produces 11.5 g PbS.

the answer is d. 74.02