Academic Subjects
➤
Chemistry
➤
General Chemistry
➤
Chemical Reactions
➤
Finding MgCl
2
Limiting Reagent
116.6 g of magnesium hydroxide is allowed to react with 500 mL of 2 M HCl.
What is the maximum amount of magnesium chloride that can be produced?
Let's start by writing a balanced equation for this reaction
Mg(OH)
2
+ 2 HCl
⇒
2 H
2
O + MgCl
2
now we'll find the moles of HCl
2 M HCl =
x moles HCl
/
0.5 L HCl
x = 1 mole HCl
We have all the information to find the limiting reagent
116.6
g Mg(OH)
2
1
mole Mg(OH)
2
/
58.33
g Mg(OH)
2
1
mole MgCl
2
/
1
mole Mg(OH)
2
95.21 g MgCl
2
/
1
mole MgCl
2
= 190.32 g MgCl
2
1
mole HCl
1
mole MgCl
2
/
2
moles HCl
95.21 g MgCl
2
/
1
mole MgCl
2
= 47.61 g MgCl
2
The limiting reagent is HCL and
the maximum amount of MgCl
2
that can be produced is 47.61 g