Sure, let's consider the given chemical reaction:
[tex]\[ \text{Mg(ClO}_3)_2 + 2 \text{NaOH} \rightarrow \text{Mg(OH)}_2 + 2\text{NaClO}_3 \][/tex]
### Step-by-Step Solution:
1. Stoichiometry: From the balanced equation, we can see the stoichiometric ratios:
- 1 mole of [tex]\(\text{Mg(ClO}_3)_2\)[/tex] reacts with 2 moles of [tex]\(\text{NaOH}\)[/tex]
- 1 mole of [tex]\(\text{Mg(ClO}_3)_2\)[/tex] produces 1 mole of [tex]\(\text{Mg(OH)}_2\)[/tex]
- 1 mole of [tex]\(\text{Mg(ClO}_3)_2\)[/tex] produces 2 moles of [tex]\(\text{NaClO}_3\)[/tex]
2. Determine the Limiting Reagent:
- Given moles of [tex]\(\text{Mg(ClO}_3)_2\)[/tex] = 2.72 moles
- Given moles of [tex]\(\text{NaOH}\)[/tex] = 3.14 moles
Since it takes 2 moles of [tex]\(\text{NaOH}\)[/tex] to react with 1 mole of [tex]\(\text{Mg(ClO}_3)_2\)[/tex], for 2.72 moles of [tex]\(\text{Mg(ClO}_3)_2\)[/tex], we would need:
[tex]\( 2 \times 2.72 = 5.44 \)[/tex] moles of [tex]\(\text{NaOH}\)[/tex].
Given only 3.14 moles of [tex]\(\text{NaOH}\)[/tex], it is clear that [tex]\(\text{NaOH}\)[/tex] is the limiting reagent.
The amount of [tex]\(\text{NaOH}\)[/tex] available should be divided by its stoichiometric coefficient (which is 2) to find out how many moles of [tex]\(\text{Mg(ClO}_3)_2\)[/tex] it can actually fully react with:
[tex]\( \frac{3.14}{2} = 1.57 \)[/tex] moles.
3. Calculate the Theoretical Yield of Products:
- The moles of [tex]\(\text{Mg(OH)}_2\)[/tex] formed will be equal to the moles of the limiting reagent (calculated above): 1.57 moles.
- The moles of [tex]\(\text{NaClO}_3\)[/tex] formed will be twice the moles of the limiting reagent: [tex]\( 2 \times 1.57 = 3.14 \)[/tex] moles.
Based on this reasoning, select the correct values from each drop-down menu.
The reaction will produce 1.57 moles of magnesium hydroxide and 3.14 moles of sodium chlorate.
