[susan__t]

 Suppose 1.00g of Mn reacts with an excess of 6.00M of HCl and I am trying to find the mass of H₂ gas produced.  Does that mean that I don't have to take account of the 6.00M HCl, and therefore just find the moles of Mn and use the ratio between Mn and H₂ to find the moles of hydrogen gas to be converted into grams??

Thank you, just seeking some clarification, I'm a little rusty on my excess/limiting reactant definitions

[Borek]

If HCl is in excess, Mn is a limiting reagent. Just find Mn moles and proceed.

[cliverlong]

Hello,

   I'm going to pretty much duplicate what Borek has written, but in a bit more detail.

   I think the phrase “substance X is in excess” makes people think the chemistry of the reaction is somehow different

I think there are two conditions the idea of “excess/limiting reagent” is trying to address

• There is “sufficient” amount of chemical X for the reaction to go to completion
• The concentration of substance X in solution does not significantly change during the reaction so this simplifies rate calculations

Let's take a specific example. Consider the reaction of calcium carbonate with hydrochloric acid. As Mitch writes, always start with a balanced equation

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)

Now, we can never exactly get the number of moles of CaCO₃ to react with the number of moles of HCl in the ratio 1:2 for both reactants to be consumed completely. So we accept that and consider one reactant is in excess so the other is consumed completely. The reactant that is consumed completely is the limiting reagent. For this example, I feel it makes more sense to think of a situation where the calcium carbonate is consumed completely – so hydrochloric acid is in excess.

How do we use this information?

If we are told hydrochloric acid is in excess then calcium carbonate is the limiting reagent and calcium carbonate is consumed completely by the reaction. First we calculate the amount of moles of calcium carbonate we are given (probably from the mass of calcium carbonate in the question). Let's say this amount is N moles. Then we use the equation to determine that 2N moles of hydrochloric acid has been consumed. We don't need any information other than “hydrochloric acid is in excess” to get the 2N value. Then we can use the equation to work out any of the other quantities produced by the reaction. The most usual would be the amount of carbon dioxide which would probably be measured as a gas volume than a mass of carbon dioxide using the approximation 1 mole of gas occupies 24dm³.

In the case of the rate, at a simplistic level (which is all I can think of at the moment) if hydrochloric acid is in excess, the concentration of hydrochloric acid does not significantly diminish through the reaction. Therefore that does not affect the rate. If we think of the CaCO₃ as a cube with sides 1cm, the initial surface area would be 6 x 1cm² = 6 cm² . Later when CaCO₃ had reacted with acid, the cube had sides 0.5cm (assume that happens) the surface area would be 6 x 0.25 cm² =  1.5 cm²  . Hence less CaCO₃ is “exposed” to the acid and the rate drops.  Eventually the reaction would stop as all the calcium carbonate would have been consumed. So you only need to consider the amount of  CaCO₃ not the acid to think about the rate.

Clive

[Borek]

The concentration of substance X in solution does not significantly change during the reaction so this simplifies rate calculations

Rate and equilibrium.

But while both these remarks are valid, they are completely irrelevant for the original question. So Susan - don't worry if you don't get this 'rate' part.