Question

1.Complete the balanced neutralization equation for the reaction below:

Answer 1

1. $H_{2}$S$O_{4}$ + Sr(OH)$_{2}$ ⇒ SrSO4 + 2 H2O is the balanced reaction.

2. 0.034 liters of KI will be required  completely react with 2.43 g of Cu(NO3)2.

3. 0.55 liters is the volume in L of a 0.724 M Nal solution contains0.405 mol of NaI.

4. 33.3 ml  many mL of 0.300 M NaF would be required to make a 0.0400 M solution of NaF when diluted to 250.0 mL with water

Explanation:

Balance chemical reaction of neutralization:

1. $H_{2}$S$O_{4}$ + Sr(OH)$_{2}$ ⇒ SrSO4 + 2 H2O

2. Data given:

balance chemical reaction:

2Cu(NO₃)₂(aq) + 4KI(aq) → 2CuI(aq) + I₂(s) + 4KNO₃(aq)

molarity of KI = 0.209 M

mass of Cu(NO₃)₂ = 2.43 grams

number of moles of Cu(NO₃)₂ will be calculated as:

number of moles = $\frac{mass}{atomic mass of 1 mole}$

atomic mass of Cu(NO₃)₂ = 187.56 grams/mole

putting the values in the equation,

number of moles= $\frac{2.43}{187.56}$

                             = 0.0129 moles

2 moles of Cu(NO₃)₂ will react with 4 moles of KI

0.0129 moles will react with x moles of KI

$\frac{4}{2}$ = $\frac{x}{0.0129}$

x = 0.0258

atomic mass of KI = 166.00 grams/mole

mass = 166.00 x 0.0258

        = 4.28 grams or ml is the final volume of KI

so molarity = $\frac{number of moles}{volume in litres}$

so molarity of KI is 0.0258 M, volume is 1 litre.

Using the formula

Minitial x Vinitial = M final Vfinal

V initial = $\frac{M final Vfinal}{Minitial}$

           = $\frac{4.28 X 0.209}{0.0258}$

            = 34.67 ml 0.034 liters of KI will be required.

3) Data given:

molarity of NaI = 0.724

number of moles of NaI =?

Volume in litres =?

formula used:

molarity = $\frac{number of moles}{volume in liters}$

volume in litres = $\frac{0.405}{0.724}$

                          = 0.55 liters is the volume

4) Data given:

Initial molarity = 0.3 M

initial volume = ?

final molarity = 0.04 M

final volume diluted by = 250 ml

formula used:

M initial X Vinitial = Mfinal X V final

putting the values in the equation:

Vinitial = $\frac{0.04 X 250}{0.3}$

            = 33.3 ml of 0.3 M solution will be required.