Answer
a. $$2PbS(s) + 3O_2(g) \longrightarrow 2PbO(s) + 2SO_2(g)$$
b. 6.02 g of $O_2$ are required to react with 30.0 g of lead (II) sulfide.
c. 17.4 g of sulfur dioxide are produced when 65.0 g of lead (II) sulfide reacts.
d. 137 g of lead (II) sulfide are used to produce 128 g of lead (II) oxide.
Work Step by Step
a.
1. Identify each reactant and product, and write the unbalanced reaction.
Reactants:
Solid lead(II) sulfide: $PbS(s)$
Oxygen gas: $O_2(g)$
Products:
Solid lead(II) oxide: $PbO(s)$
Sulfur dioxide gas: $SO_2(g)$
$$PbS(s) + O_2(g) \longrightarrow PbO(s) + SO_2(g)$$
2. Balance the equation.
- There are 3 oxygens in the products side, but only 2 one the reactants side. In order to add an extra oxygen atom to the reactants side, we can put $\frac 32$ as the coefficient of $O_2(g)$.
$$PbS(s) + \frac 32O_2(g) \longrightarrow PbO(s) + SO_2(g)$$
Now, everything is balanced. We only need to remove the fraction, by multiplying all coefficients by 2:
$$2PbS(s) + 3O_2(g) \longrightarrow 2PbO(s) + 2SO_2(g)$$
For b., c. and d.:
- Find the conversion factors and calculate each mass.
b. $$ \frac{ 2 \space moles \space PbS }{ 3 \space moles \space O_2 } \space and \space \frac{ 3 \space moles \space O_2 }{ 2 \space moles \space PbS }$$
$ PbS $ : ( 207.2 $\times$ 1 )+ ( 32.06 $\times$ 1 )= 239.3 g/mol
$$ \frac{1 \space mole \space PbS }{ 239.3 \space g \space PbS } \space and \space \frac{ 239.3 \space g \space PbS }{1 \space mole \space PbS }$$
$ O_2 $ : ( 16.00 $\times$ 2 )= 32.00 g/mol
$$ \frac{1 \space mole \space O_2 }{ 32.00 \space g \space O_2 } \space and \space \frac{ 32.00 \space g \space O_2 }{1 \space mole \space O_2 }$$
$$ 30.0 \space g \space PbS \times \frac{1 \space mole \space PbS }{ 239.3 \space g \space PbS } \times \frac{ 3 \space moles \space O_2 }{ 2 \space moles \space PbS } \times \frac{ 32.00 \space g \space O_2 }{1 \space mole \space O_2 } = 6.02 \space g \space O_2 $$
c. $$ \frac{ 2 \space moles \space PbS }{ 2 \space moles \space SO_2 } \space and \space \frac{ 2 \space moles \space SO_2 }{ 2 \space moles \space PbS }$$
$ PbS $ : ( 207.2 $\times$ 1 )+ ( 32.06 $\times$ 1 )= 239.3 g/mol
$$ \frac{1 \space mole \space PbS }{ 239.3 \space g \space PbS } \space and \space \frac{ 239.3 \space g \space PbS }{1 \space mole \space PbS }$$
$ SO_2 $ : ( 16.00 $\times$ 2 )+ ( 32.06 $\times$ 1 )= 64.06 g/mol
$$ \frac{1 \space mole \space SO_2 }{ 64.06 \space g \space SO_2 } \space and \space \frac{ 64.06 \space g \space SO_2 }{1 \space mole \space SO_2 }$$
$$ 65.0 \space g \space PbS \times \frac{1 \space mole \space PbS }{ 239.3 \space g \space PbS } \times \frac{ 2 \space moles \space SO_2 }{ 2 \space moles \space PbS } \times \frac{ 64.06 \space g \space SO_2 }{1 \space mole \space SO_2 } = 17.4 \space g \space SO_2 $$
d. $$ \frac{ 2 \space moles \space PbO }{ 2 \space moles \space PbS } \space and \space \frac{ 2 \space moles \space PbS }{ 2 \space moles \space PbO }$$
$ PbO $ : ( 207.2 $\times$ 1 )+ ( 16.00 $\times$ 1 )= 223.2 g/mol
$$ \frac{1 \space mole \space PbO }{ 223.2 \space g \space PbO } \space and \space \frac{ 223.2 \space g \space PbO }{1 \space mole \space PbO }$$
$ PbS $ : ( 207.2 $\times$ 1 )+ ( 32.06 $\times$ 1 )= 239.3 g/mol
$$ \frac{1 \space mole \space PbS }{ 239.3 \space g \space PbS } \space and \space \frac{ 239.3 \space g \space PbS }{1 \space mole \space PbS }$$
$$ 128 \space g \space PbO \times \frac{1 \space mole \space PbO }{ 223.2 \space g \space PbO } \times \frac{ 2 \space moles \space PbS }{ 2 \space moles \space PbO } \times \frac{ 239.3 \space g \space PbS }{1 \space mole \space PbS } = 137 \space g \space PbS $$
