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A reaction proceeds with 2.72 moles of magnesium chlorate and 3.14 moles of sodium hydroxide. This is the equation of the reaction:
Mg(ClO3)2 + 2NaOH → Mg(OH)2 + 2NaClO3.

A reaction proceeds with 272 moles of magnesium chlorate and 314 moles of sodium hydroxide This is the equation of the reaction MgClO32 2NaOH MgOH2 2NaClO3 class=

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jcherry99

Mg(ClO3)2 + 2NaOH → Mg(OH)2 + 2NaClO3.

This is about as hard a question as you can get in beginning chemistry.  You have to figure out which chemical between the two reactants is the LEAST number of moles before you can do anything about the products. Read that sentence over again carefully and keep in mind what least means. It's the key to the entire question.

For every mole of Mg(ClO3)2 you need two moles of NaOH. That's what the balanced equation tells you. The big number on the left of the chemical is where I'm getting 1 to 2. One is in front of Mg(ClO3)2 and 2 is to the left of NaOH

So you have 2.72 moles of Mg(ClO3)2. You would need to have 2.72*2 moles of NaOH which is 5.44 moles of NaOH. Do you have enough NaOH. (I should hear a resounding NO.) So the limiting reactant is NaOH. You only have 3.14 moles of NaOH. Now you can start getting the answer to the question

You now need to set up the same ratio for the Mg(OH)2

For every 2 moles of NaOH you get 1 mole of Mg(OH)2. The proportion is

NaOH/Mg(OH)2 = 2/1

NaOH = 3.14 moles

3.14 / Mg(OH)2 = 2/1     Cross multiply

3.14 = 2 * Mg(OH)2       Divide by 2

3.14/2 = Mg(OH)2          Combine the left

1.57 = Mg(OH)2              Answer  

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I'll let you work through the second blank. I'm not completely heartless so I'll tell you that the ratio is 1:1 and that the answer should be 3.14

============

Answer Box one = 1.57

Answer Box Two = 3.14


znk

Answer:

1.57 mol; 3.14 mol

Explanation:

We have the amounts of two reactants, so this is a limiting reactant problem.  

We know that we will need a balanced equation with moles of the compounds involved.  

Step 1. Gather all the information in one place.

           Mg(ClO₃)₂ + 2NaOH ⟶ Mg(OH)₂ + 2NaClO₃

n/mol:      2.72             3.14

Step 2. Identify the limiting reactant

Calculate the moles of Mg(OH)₂ we can obtain from each reactant.  

From Mg(ClO₃)₂: The molar ratio of Mg(OH)₂: Mg(ClO₃)₂ is 1:1 .

Moles of Mg(OH)₂ = 2.72 × 1/1

Moles of Mg(OH)₂ = 2.72mol Mg(OH)₂

From NaOH: The molar ratio of Mg(OH)₂:NaOH is 1:2.

Moles of Mg(OH)₂ = 3.14 × 1/2

Moles of Mg(OH)₂ = 1.57 mol Mg(OH)₂

NaOH is the limiting reactant because it gives the smaller amount of Mg(OH)₂.

Step 3. Calculate the moles of NaClO₃.

The molar ratio of NaClO₃:NaOH is 2:2.

Moles of NaClO₃ = 3.14 × 2/2

Moles of NaClO₃ = 3.14  mol NaClO₃

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