#excess reactant

[jessicaauendoeschemistry-blog]

Metallic magnesium reacts with steam to produce magnesium hydroxide and hydrogen gas.If 16.3 g of Mg are heated with 8 g of H2O,what is the limiting reactant?1. Mg2. H2OHow much of the excess reactant is left?Answer in units of mol.How much Mg(OH)2 is formed?Answer in units of g.How much H2 is formed?Answer in units of g.--can you please explain this, i have my midterm Wednesday &i really need to learn how to do these >.<

The first step, as with all chemistry problems, is to balance your equation.

Then you plug in your given values for each reactant to find how much product is formed. In this step, you can use any product you want. I chose Mg(OH)2 for no particular reason. Either way, it will give you the same result!

Limiting reactant means "which reactant will run out first?" So there will almost always be an excess reactant. In this case, the excess is H2O. We can find how much remains with a simple stoichiometric equation. Start with your limiting reactant and use a mole-to-mole ratio to find how much excess is used. Then subtract that amount from how much you had to begin with. Sorry about the scan here, the final answer is 10.9 grams. I see now that you wanted that in moles, so you can just divide this number by MM (molar mass in grams/mol). The final answer is .448mol remaining.

You use the same stoichiometric equation to find how much of each product is formed.

And there you go! Good luck on your midterm!

[jessicaauendoeschemistry-blog]

Some Help With Your Limiting Reactant Question

lilcheeriox3:

A mixture of 11.297 g if iron(II) oxide and 5.773 g of aluminum metal is placed in a crucible and heated in a high temperature oven, where a reduction of the oxide takes place.

3 FeO(s) + 2Al(ℓ) → 3 Fe(ℓ) + Al2O3(s) 

(I know the limiting reactant is FeO) From there I don’t know what else to do.

+DIRECTIONS:

1. Find the maximum amount of iron that can be produced.  2. Then, calculate the mass of excess reactant remaining in the crucible.

Answer in units of g.

If you could help, it means so much. Thank you btw! :) thank youu so much btw.

Oh, & this is what steps I have so far & please check if my calculations are correct:

3 FeO(s) + 2Al(ℓ) → 3 Fe(ℓ) + Al2O3(s)

1a.) 11.297g.FeO (1mol.FeO / 71.85g.FeO) = 0.157230341 mol.FeO

1b.) 5.773g.Al (1mol. Al / 26.98g. Al) = 0.2139733136 mol. Al

2a.) 0.157230341 mol.FeO (3 mol.Fe / 3 mol. FeO) = 0.157230341 mol.Fe

2b.) 0.2139733136 mol. Al (3 mol. Fe / 2 mol. Al) = 0.3209599703 mol. Al

(*Therefore, FeO is the limiting reactant)

3.) 0.157230341 mol.Fe (55.85g. Fe /1 mol. Fe) = 8.781314544g. Fe

Apparantly that was wrong :// I don’t know what I did wrong. Please, I need help ):

Don't fret! Limiting reactant problems are some of the most popular in gen chem because they test you on so many concepts at once: balancing equations, stoichiometry, limiting reactant, excess reactant... I worked through the problem with you and found the same mass of iron produced. I recommend not trusting the back of your book or your answer sheets completely; in my experience, they are wrong more often than one would like.

Here is what I did:

Hope this helps!