limiting reactant with two products???

Question

ok...5.87g of Mg(OH)2 reacts with 12.84g of HCl to form MgCl2 and water. I have the balanced equation but im not sure what to do after that..Please help THANKS!!

thanks for the help!!

Answer 1

Here is a link that will explain it to you thoroughly:

http://members.aol.com/profchm/limits.html

or

http://en.wikipedia.org/wiki/Limiting_reagent

hope that helps!

Answer 2

you advise if u react them at the same time? you should write out an equation and stability it then artwork out the mol of all the reactants and divide the mol via the extensive style infront of the compound interior the eqaution. the proscribing reactant is the smaller mol. isn't achieveable to have 2 proscribing reactants. for all of them for use up they should have the the excellent option mass to make the the excellent option mol which will react completly with the the excellent option ratio

Answer 3

To find the limiting reactant:
Steps:
1. Take one reactant convert to moles and use the mole ratio to get to moles of water.

2. Take the other reactant, convert to moles and use the mole ratio to get to moles of water.

3. Compare the mole calculations, the reactant that produced the least amount of moles of water is the limiting reactant.

Answer 4

Mg(OH)2+2HCL-------------->MgCl2+H2O

No. of Mole of Mg(OH)2=5.87\58=0.1012 mol
No. of Mole of HCl= 12.84\36.5=0.351

Mg(OH)2= .1012\1= 0.1012 mol
HCL=.351\2=0.175 mol

So Mg(oH)2 is the LR

1 mole of Mg(OH)2------------>1 mole of MgCl2
0.1012 Mole of Mg(OH)2-------->?? mole of MgCl2

No of mole of MgCl2= 0.1012 mole
Weight of MgCl2= 0.1012*95=9.614 gm and this is called theoretical yeild.