The reaction forms 11.5 g PbS.
We have the masses of two reactants, so this is a <em>limiting reactant</em> problem.
We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.
<em>Step 1</em>. <em>Gather all the information</em> 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
<em>Step 2</em>. Calculate the <em>moles of each reactant </em>
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
S<em>tep 3</em>. Identify the <em>limiting reactant</em>
Calculate the moles of PbS we can obtain from each reactant.
<em>From Pb</em>: Moles of PbS = 0.048 26 mol Pb × (1 mol PbS/1 mol Pb )
= 0.048 26 mol PbS
<em>From S</em>: Moles of S = 0.9357 mol S × (1 mol PbS/1 mol S) = 0.9357 mol PbS
<em>Pb is the limiting reactant</em> because it gives the smaller amount of PbS.
<em>Step 4</em>. Calculate the <em>mass of PbS</em>.
Mass = 0.048 26 mol PbS × (239.28 g PbS/1 mol PbS) = 11.5 g PbS
The reaction produces 11.5 g PbS.
