Answered

Welcome to Westonci.ca, where finding answers to your questions is made simple by our community of experts. Ask your questions and receive detailed answers from professionals with extensive experience in various fields. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

Select the correct answer from each drop-down menu.

A reaction proceeds with 2.72 moles of magnesium chlorate and 3.14 moles of sodium hydroxide. This is the equation:
[tex]Mg(ClO_3)_2 + 2 NaOH \rightarrow Mg(OH)_2 + 2 NaClO_3[/tex]

Determine the theoretical amount of each product that the reaction will produce.

The reaction will produce [tex]$\square$[/tex] moles of magnesium hydroxide and [tex]$\square$[/tex] moles of sodium chlorate.

Sagot :

GinnyAnswer
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.
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.