Answer
The drawing of the product mixture must have 6 $SO_3$ molecules and 3 $O_2$ molecules.
$SO_2$ is the limiting reactant.
If 96.0 g of $SO_2$ react with 32.0 g of $O_2$, 120. g of $SO_3$ will be formed.
Work Step by Step
1. Count each amount of molecules:
6 $SO_2$ and 6 $O_2$:
2. Write the balanced reaction:
$$2 \ SO_2 + O_2 \longrightarrow 2 \ SO_3$$
3. Therefore, 6 molecules of $SO_2$ reacts with 12 $O_2$:
$O_2$ excess.
$SO_2$ limiting.
4. Now, knowing the limiting reactant, we can calculate the amount of $SO_3$ formed.
Each 2 $SO_2$ forms 2 $SO_3$. Thus, 6 $SO_2$ produces 6 $SO_3$.
At the same time, 3 $O_2$ will be consumed in the reaction, and 3 will remain in the products.
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- Calculate or find the molar mass for $ SO_2 $:
$ SO_2 $ : ( 16.00 $\times$ 2 )+ ( 32.07 $\times$ 1 )= 64.07 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 96.0 \space g \times \frac{1 \space mole}{ 64.07 \space g} = 1.50 \space moles$$
- Calculate or find the molar mass for $ O_2 $:
$ O_2 $ : ( 16.00 $\times$ 2 )= 32.00 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 32.0 \space g \times \frac{1 \space mole}{ 32.00 \space g} = 1.00 \space mole$$
Find the amount of product if each reactant is completely consumed.
$$ 1.50 \space moles \space SO_2 \times \frac{ 2 \space moles \ SO_3 }{ 2 \space moles \space SO_2 } = 1.50 \space moles \space SO_3 $$
$$ 1.00 \space mole \space O_2 \times \frac{ 2 \space moles \ SO_3 }{ 1 \space mole \space O_2 } = 2.00 \space moles \space SO_3 $$
Since the reaction of $ SO_2 $ produces less $ SO_3 $ for these quantities, it is the limiting reactant.
- Calculate or find the molar mass for $ SO_3 $:
$ SO_3 $ : ( 16.00 $\times$ 3 )+ ( 32.07 $\times$ 1 )= 80.07 g/mol
- Using the molar mass as a conversion factor, find the mass in g:
$$ 1.50 \space mole \times \frac{ 80.07 \space g}{1 \space mole} = 120. \space g$$
