Question

Q1. a) If 26.4 mL of a solution of NaOH are used to react 0.561 g of oxalic acid dihydrate (C2O4H2•2H2O) according to the following equation, what molarity is the NaOH solution? 2 NaOH (aq) + C2O4H2•2H2O (s)  C2O4Na2 (aq) + 4 H2O (l)

b) If this NaOH solution was formed as a dilute solution from an original solution containing 60.0 g NaOH in 0.500 L solution, how many mL of the original solution were used to make 150. mL of the dilute solution?

Q 2. 2. a) How many grams of lithium nitrate are produced from the reaction of 70.0 mL of a 2.186 x 10-2 M solution of lead(II) nitrate and 80.0 mL of a 2.143 x 10-2 M solution of lithium sulfate?

b) What is the limiting reactant?

Answer 1

1.a) The given reaction is: 2 NaOH (aq) + C₂O₄H₂•2H₂O (s) C₂O₄Na₂ (aq) + 4 H₂O (l)

Mass of the oxalic acid dihydrate = 0.561 g

Molar mass of the oxalic acid dihydrate = 126 g/mol

Number of moles of  oxalic acid dihydrate = mass/molar mass = 0.561 g/126 g/mol = .0045 mole

From the reaction, it is observed that 1 mole of oxalic acid dihydrate reacts with NaOH = 2 moles

Hence, 0.0045 mole of oxalic acid dihydrate reacts with NaOH = 2 X 0.0045 = 0.009 mole

Volume of NaOH solution = 26.4 mL = 0.0264 L

Hence, molarity of NaOH solution = number of moles / volume of solution in L = 0.009 mole / 0.0264 L

= 0.341 mole/L = 0.341 M

b) Mass of NaOH = 60 g

Molar mass of NaOH = 40 g/mol

Number of moles = 60 g/40 g/mol = 1.5 mol

Volume of the stock solution = 0.500 L

Molarity of the stock solutiuon = 1.5 mol/0.500 L = 3 M

Here, M₁ = 3, V₁=? and M₂ = 0.341 M , V₂ = 150 mL

Applying M₁V₁ = M₂V₂   or, V₁ = M₂V₂/M₁ = 0.341 M X 150 mL / 3 M = 17.05 mL