Answer to Question #248672 in General Chemistry for STURAH

Question

4. An aqueous solution of Mg(OH)2 is added to an aqueous solution of HCl to form solid  MgCl2 and H2O. If 40.5g of Mg(OH)2 and 35.0g of HCl are used: 

a) Define limiting reactant 

b) Determine the limiting reactant. 

c) Calculate the mass of MgCl2 produced after the reaction has completed. d) If the percentage yield of MgCl2 is 85%, calculate the actual mass of MgCl2  obtained. 

[ Ar : Mg =24.3 , H=1.0 , O = 16.0 , Cl = 35.5 ] 

Answer 1

a) The reactant that controls the amount of product able to be produced by a chemical reaction because it is used up completely.

b) Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O

M(Mg(OH)₂) = 58.3 g\mol

M(HCl) = 36.5 g/mol

n(Mg(OH)₂) $= \frac{40.5}{58.3} = 0.694 \;mol$

n(HCl) $= \frac{35.0}{36.5} = 0.959 \;mol$

According to the reaction for one mole of Mg(OH)₂ we need two moles of HCl. So, HCl is limitting reactant.

c) n(MgCl₂) $= \frac{1}{2}n(HCl) = \frac{1}{2} \times 0.959 = 0.479 \;mol$

M(MgCl₂) = 95.3 g/mol

m(MgCl₂) $= 95.3 \times 0.479 = 45.65 \;g$

d) Proportion:

45.65 g – 100 %

x g – 85 %

$x = \frac{45.65 \times 85}{100} = 38.80 \;g$