Moles of Mg(OH)2, HNO3, Mg(NO3)2 in Reaction

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In summary, After adding 5.53g of Mg(OH)2 to 25.0mL of 0.200M HNO3, the balanced reaction equation is Mg(OH)2 + 2HNO3 -> Mg(NO3)2 + 2H2O. The limiter in this case is HNO3, and there is 0.315g of HNO3 in the solution. Using stoichiometry, it can be determined that there will be 0 moles of HNO3 left after the reaction is complete. The number of moles of Mg(OH)2 will be determined by converting the mass of 5.53g to moles. According to the coefficients
  • #1
MellowOne
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Homework Statement


A sample of 5.53g of Mg(OH)2 is added to 25.0mL of 0.200M HNO3. How many moles of Mg(OH)2, HNO3, and Mg(NO3)2 are present after the reaction is complete?

Homework Equations


M = mol/L


The Attempt at a Solution


The balanced equation for the reaction is Mg(OH)2 + 2HNO3 -> Mg(NO3)2 + 2H2O

The limiter in this case would be HNO3 and I found that in 0.025L of this solution there is 0.315g of HNO3.

What I'm not sure is how to find the moles of each after the reaction is complete. I assume it's with stoichiometry, but I'm not sure how to find the moles of each at the end. I know that there should be 0 moles of HNO3 because it's used up correct? As for the Mg(OH)2, am I suppose to subtract the grams of HNO3 from the 5.53g of Mg(OH)2?
 
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  • #2
MellowOne said:
I know that there should be 0 moles of HNO3 because it's used up correct?

That's right.

Convert Mg(OH)2 to moles.

Look at the coefficients in the reaction equation.

How many moles of Mg(OH)2 will react with HNO3?

How many moles of MgCl2 will be produced?

How many moles of Mg(OH)2 will be left?
 
  • #3



I would first convert the given mass of Mg(OH)2 to moles by using its molar mass (58.33 g/mol). This gives us 0.095 moles of Mg(OH)2. Next, using the given volume and molarity of HNO3, we can calculate the initial moles of HNO3 (0.005 moles).

Since HNO3 is the limiting reagent, it will be completely consumed in the reaction. This means that there will be no HNO3 left after the reaction is complete.

To find the moles of Mg(NO3)2 produced, we can use the mole ratio from the balanced equation. For every 1 mole of Mg(OH)2, we get 1 mole of Mg(NO3)2. Therefore, the moles of Mg(NO3)2 produced will also be 0.095 moles.

Finally, to find the moles of H2O produced, we can use the same mole ratio as before. For every 1 mole of Mg(OH)2, we get 2 moles of H2O. This means that 0.095 moles of Mg(OH)2 will produce 0.190 moles of H2O.

In summary, after the reaction is complete, there will be 0 moles of HNO3, 0.095 moles of Mg(NO3)2, and 0.190 moles of H2O present. It is important to note that these are the theoretical values and in a real experiment, there may be slight variations due to experimental error.
 

1. What is the mole ratio between Mg(OH)2 and HNO3 in the reaction?

The mole ratio between Mg(OH)2 and HNO3 in the reaction is 1:2. This means that for every 1 mole of Mg(OH)2, 2 moles of HNO3 are required for the reaction to take place.

2. How many moles of Mg(NO3)2 will be produced when 0.5 moles of Mg(OH)2 react?

Using the mole ratio of 1:1 between Mg(OH)2 and Mg(NO3)2, we can calculate that 0.5 moles of Mg(OH)2 will produce 0.5 moles of Mg(NO3)2.

3. How many moles of HNO3 are needed to react with 1 mole of Mg(OH)2?

As mentioned earlier, the mole ratio between Mg(OH)2 and HNO3 is 1:2. Therefore, 1 mole of Mg(OH)2 will require 2 moles of HNO3 for the reaction to occur.

4. What is the molar mass of Mg(NO3)2?

The molar mass of Mg(NO3)2 is 148.31 g/mol. This can be calculated by adding the molar masses of each element present in the compound (Mg = 24.31 g/mol, N = 14.01 g/mol, O = 16.00 g/mol).

5. What is the limiting reactant in this reaction?

The limiting reactant in this reaction is the reactant that is completely consumed and limits the amount of product that can be formed. To determine the limiting reactant, you would need to compare the number of moles of each reactant present and see which one is present in a lower amount compared to the mole ratio. The reactant with the lower amount will be the limiting reactant.

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