Question

Imagine you have two solutions, one that is 0.113 M in HCl and the other that is 0.085 M in Sr(OH)2.

If you add 37.4 mL of the HCl solution to 31.8 mL of the Sr(OH)2 solution, which is the limiting reactant?

How many grams of SrCl2 will be formed?

How many moles of the excess reagent remain in solution?

Answer 1

Sr(OH)₂ + HCl SrCl₂ + 2H₂O

The balanced equation is expressed as

Sr(OH)₂ + 2HCl    SrCl₂ + 2H₂O

Molarity = Moles / Liter

Moles = Molarity x Liter

Now,

37.4 mL of 0.113 M HCl

Moles of HCl = 0.113 M x 0.0374 L = 0.004 moles

31.8 mL of 0.085 M Sr(OH)₂

Moles of Sr(OH)₂ = 0.085 M x 0.0318 L = 0.003 moles

Now,

Sr(OH)₂ + 2HCl    SrCl₂ + 2H₂O

In the above reaction equation

1 mole of Sr(OH)₂ reacts with 2 moles of HCl to produce 1 mole of SrCl₂ and 2 moles of H₂O.

Since moles of HCl is 0.004 mol (less than 0.003/2 moles of Sr(OH)₂. So, HCl is limiting reagent and Sr(OH)₂ is taken excess.

Now, the excess reagent = 0.003 mol – 0.002 mol = 0.001 mol

Now,

Sr(OH)₂ + 2HCl    SrCl₂ + 2H₂O

In the above reaction equation

2 moles of HCl produces 1 mole of SrCl₂.

1 moles of HCl produces 1/2 mole of SrCl₂.

0.004 moles of HCl produces 0.004/2 mole of SrCl₂.

0.004 moles of HCl produces 0.002 mole of SrCl₂.