Which one does the MOLES of product depend on, is the "LIMITED" or "EXCESS" reagent?

So I completely understand the whole concept of finding the limited and excess reagent, but which one does the product depend on?

In a displacement reaction (measuring Enthalpy Change), when calculating the moles of the reactants, you'd choose the limited reagent.

But, why would they use the excess reagent when calculating the moles of the product, like in this case :s-smilie:

??

Original post by Adorable98

So I completely understand the whole concept of finding the limited and excess reagent, but which one does the product depend on?

In a displacement reaction (measuring Enthalpy Change), when calculating the moles of the reactants, you'd choose the limited reagent.

But, why would they use the excess reagent when calculating the moles of the product, like in this case :s-smilie:

??

The limiting reagent is the one which is used up first. It limits the extent of the reactions which can then not proceed.

Hence the limiting reagent determines the product amounts.

In the example above all of the HCl is used up. The product moles (using the stoichiometry of the equation) = mol HCl/2

(edited 8 years ago)

Reply 2

Adorable98

OP

15

Original post by charco

The limiting reagent is the one which is used up first. It limits the extent of the reactions which can then not proceed.

Hence the limiting reagent determines the product amounts.

I see, but why did the question (pic I posted above) use the excess reagent (4.77 in this case) instead of the limiting reagent (13.5)? :smile:

Original post by Adorable98

I see, but why did the question (pic I posted above) use the excess reagent (4.77 in this case) instead of the limiting reagent (13.5)? :smile:

It didn't.

The magnesium is in excess.

There are 6.75 mol initially and only 4.77 moles are used in the reaction.

Reply 4

Adorable98

OP

15

Original post by charco

It didn't.

The magnesium is in excess.

There are 6.75 mol initially and only 4.77 moles are used in the reaction.

So they did use the excess reagent?! :s-smilie:

Magnesium is in excess.
we have 6.75 moles of Magnesium, but only need 4.77 moles, so yes it's in excess. but as you could see at the left hand side, 4.77 is from the magnesium? Am I not right? :s-smilie:

Original post by Adorable98

So they did use the excess reagent?! :s-smilie:

Magnesium is in excess.
we have 6.75 moles of Magnesium, but only need 4.77 moles, so yes it's in excess. but as you could see at the left hand side, 4.77 is from the magnesium? Am I not right? :s-smilie:

No.

At the start of the reaction you are told that there are 6.75 mol of Mg and 9.54 mol of HCl

Using this information you cannot get the moles of products WITHOUT determining the limiting reagent.

The limiting reagent is HCl.

It is this that then determines how many moles of Mg can react. This is also the moles of the two products.

Reply 6

Adorable98

OP

15

Original post by charco

No.

At the start of the reaction you are told that there are 6.75 mol of Mg and 9.54 mol of HCl

Using this information you cannot get the moles of products WITHOUT determining the limiting reagent.

The limiting reagent is HCl.

It is this that then determines how many moles of Mg can react. This is also the moles of the two products.

Okay, so I understand that the HCL is the limiting reagent whereas the Mg is the one in excess.
And you told me that the moles of the product, 4.77 moles is NOT from the Magnesium.

So how did they calculate the moles of the products and end up getting 4.77? That's the part that I'm not still getting? :s-smilie:

Reply 7

Adorable98

OP

15

Bump! Bump! Bump! :sad:

Original post by Adorable98

Okay, so I understand that the HCL is the limiting reagent whereas the Mg is the one in excess.
And you told me that the moles of the product, 4.77 moles is NOT from the Magnesium.

So how did they calculate the moles of the products and end up getting 4.77? That's the part that I'm not still getting? :s-smilie:

The equation for the reaction is:

Mg + 2HCl -> MgCl₂ + H₂

This says that:

1 mol of magnesium must react with 2 moles of HCl to produce 1 mol of magnesium chloride and 1 mol of hydrogen.

In other words you must divide the moles of HCl reacting by 2 to get the moles of products.

If you have 9.54 mol of HCl you must get 9.54/2 mol of magnesium chloride and hydrogen produced

9.54/2 = 4.77

Reply 9

Adorable98

OP

15

Original post by charco

The equation for the reaction is:

Mg + 2HCl -> MgCl₂ + H₂

This says that:

1 mol of magnesium must react with 2 moles of HCl to produce 1 mol of magnesium chloride and 1 mol of hydrogen.

In other words you must divide the moles of HCl reacting by 2 to get the moles of products.

If you have 9.54 mol of HCl you must get 9.54/2 mol of magnesium chloride and hydrogen produced

9.54/2 = 4.77

I see, so you first find the limiting reagent, then divide it according to the ratio!! Thaanx!! :h:

Original post by Adorable98

I see, so you first find the limiting reagent, then divide it according to the ratio!! Thaanx!! :h:

corrrrrrrrrrrect ....

Reply 11

Infraspecies

16

If I have 34 people and 40 hats, how many fully hatted people may I have?

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Reply 12

Adorable98

OP

15

Original post by Infraspecies

If I have 34 people and 40 hats, how many fully hatted people may I have?

If I have 48 people and 94 achiral gloves, how many fully gloved people may I have?