Answer
a. $$2H_2S(g) + 3O_2(g) \longrightarrow 2SO_2(g) + 2H_2O(g)$$
b. 3.52 g of $O_2$
c. 51.4 g of $SO_2$
d. 149 g of $O_2$
Work Step by Step
a.
1. Identify each reactant and product, and write the unbalanced reaction.
Reactants:
Dihydrogen sulfide gas: $H_2S(g)$
Oxygen gas: $O_2(g)$
Products:
Sulfur dioxide gas: $SO_2(g)$
Water vapor: $H_2O(g)$
$$H_2S(g) + O_2(g) \longrightarrow SO_2(g) + H_2O(g)$$
2. Balance the equation.
- There are 3 oxygens in the products side, but only 2 on the reactants side. In order to add an extra oxygen atom to the reactants side, we can put $\frac 32$ as the coefficient of $O_2(g)$.
$$H_2S(g) + \frac 32O_2(g) \longrightarrow SO_2(g) + H_2O(g)$$
Now, everything is balanced. We only need to remove the fraction, by multiplying all coefficients by 2:
$$2H_2S(g) + 3O_2(g) \longrightarrow 2SO_2(g) + 2H_2O(g)$$
For b., c. and d.:
- Find the conversion factors and calculate each mass.
b. $$ \frac{ 2 \space moles \space H_2S }{ 3 \space moles \space O_2 } \space and \space \frac{ 3 \space moles \space O_2 }{ 2 \space moles \space H_2S }$$
$ H_2S $ : ( 1.008 $\times$ 2 )+ ( 32.06 $\times$ 1 )= 34.08 g/mol
$$ \frac{1 \space mole \space H_2S }{ 34.08 \space g \space H_2S } \space and \space \frac{ 34.08 \space g \space H_2S }{1 \space mole \space H_2S }$$
$ O_2 $ : ( 16.00 $\times$ 2 )= 32.00 g/mol
$$ \frac{1 \space mole \space O_2 }{ 32.00 \space g \space O_2 } \space and \space \frac{ 32.00 \space g \space O_2 }{1 \space mole \space O_2 }$$
$$ 2.50 \space g \space H_2S \times \frac{1 \space mole \space H_2S }{ 34.08 \space g \space H_2S } \times \frac{ 3 \space moles \space O_2 }{ 2 \space moles \space H_2S } \times \frac{ 32.00 \space g \space O_2 }{1 \space mole \space O_2 } = 3.52 \space g \space O_2 $$
c. $$ \frac{ 3 \space moles \space O_2 }{ 2 \space moles \space SO_2 } \space and \space \frac{ 2 \space moles \space SO_2 }{ 3 \space moles \space O_2 }$$
$ O_2 $ : ( 16.00 $\times$ 2 )= 32.00 g/mol
$$ \frac{1 \space mole \space O_2 }{ 32.00 \space g \space O_2 } \space and \space \frac{ 32.00 \space g \space O_2 }{1 \space mole \space O_2 }$$
$ SO_2 $ : ( 16.00 $\times$ 2 )+ ( 32.06 $\times$ 1 )= 64.06 g/mol
$$ \frac{1 \space mole \space SO_2 }{ 64.06 \space g \space SO_2 } \space and \space \frac{ 64.06 \space g \space SO_2 }{1 \space mole \space SO_2 }$$
$$ 38.5 \space g \space O_2 \times \frac{1 \space mole \space O_2 }{ 32.00 \space g \space O_2 } \times \frac{ 2 \space moles \space SO_2 }{ 3 \space moles \space O_2 } \times \frac{ 64.06 \space g \space SO_2 }{1 \space mole \space SO_2 } = 51.4 \space g \space SO_2 $$
d. $$ \frac{ 2 \space moles \space H_2O }{ 3 \space moles \space O_2 } \space and \space \frac{ 3 \space moles \space O_2 }{ 2 \space moles \space H_2O }$$
$ H_2O $ : ( 1.008 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 18.02 g/mol
$$ \frac{1 \space mole \space H_2O }{ 18.02 \space g \space H_2O } \space and \space \frac{ 18.02 \space g \space H_2O }{1 \space mole \space H_2O }$$
$ O_2 $ : ( 16.00 $\times$ 2 )= 32.00 g/mol
$$ \frac{1 \space mole \space O_2 }{ 32.00 \space g \space O_2 } \space and \space \frac{ 32.00 \space g \space O_2 }{1 \space mole \space O_2 }$$
$$ 55.8 \space g \space H_2O \times \frac{1 \space mole \space H_2O }{ 18.02 \space g \space H_2O } \times \frac{ 3 \space moles \space O_2 }{ 2 \space moles \space H_2O } \times \frac{ 32.00 \space g \space O_2 }{1 \space mole \space O_2 } = 149 \space g \space O_2 $$
