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bixtya [17]
3 years ago
9

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

Chemistry
1 answer:
zaharov [31]3 years ago
4 0

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.

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