Question

Nitric acid and magnesium hydroxide react to form magnesium nitrate and water

A) how many moles are in 83.1 ml of 1.581M nitric acid?

B) how many moles r in 54.3ml of 1.815M magnesium hydroxide?

C) what is the limiting reactant, and what is the theoretical yield of magnesium nitrate in gram?

D) if 2.831 g of mag. Nitrate r produced in the reaction what is the percent yield

Answer 1

(A) Given Molarity of HNO₃(nitric acid) , M = 1.581 M

                                 Volume of HNO₃ is , V = 83.1 mL = 0.0831 L

Number of moles of HNO₃ is , n = Molarity x volume in L

                                                = 1.581 M x 0.0831 L

                                               = 0.131 moles

Therefore the number of moles of HNO₃ is 0.131 mol

(B) Number of moles of Magnesium hydroxide,Mg(OH)₂ is , n = Molarity x volume in L

                                               = 1.815 M x 0.0543 L

                                                 = 0.098 moles

Therefore the number of moles of Mg(OH)₂ is 0.098 mol

(C) The balanced reaction when Nitric acid and magnesium hydroxide react to form magnesium nitrate and water is

2HNO₃ + Mg(OH)₂ Mg(NO₃)₂ + 2H₂O

According to the balanced equation ,

2 moles of HNO₃ reacts with 1 mole of Mg(OH)₂

0.131 moles of HNO₃ reacts with M mole of Mg(OH)₂

M = ( 0.131x1) / 2

    = 0.065 moles

So 0.098 - 0.065 = 0.033 moles of Mg(OH)₂ left unreacted so Mg(OH)₂   is the excess reactant.

Since all the mass of HNO₃ completly reacted , so HNO₃ is the limiting reactant.

(D) From the balanced reaction ,

2 moles of HNO₃ produces 1 mole of Mg(NO₃)₂

0.131 moles of HNO₃ produces N mole of Mg(NO₃)₂

N = (0.131x1)/2

   = 0.0655 moles

Molar mass of Mg(NO₃)₂ = 24.3+2[(14+(3x16)]

                                     = 148.3 g/mol

So mass of Mg(NO₃)₂ produced is , m = molar mass xnumber of moles

                                                        = 148.3(g/mol) x 0.0655 mol

                                                        = 9.714 g    ----> which is the theoretical yield

& yield = ( actual yield/ theoretical yield) x100

          = ( 2.831/ 9.714)x100

         = 29.1 %