Question

1.Write the balanced molecular and net ionic equations for the reaction between aluminum metal and silver nitrate. Identify the oxidation and reduction half-reactions.

Balanced molecular equation:

Net ionic equation:

Oxidation half-reaction:

Reduction half-reaction:

2.According to the label on a bottle of concentrated hydrochloric acid, the contents are 36.0% HCl by mass and have a density of 1.18 g/mL.

What is the molarity of concentrated HCl?

What volume of it would you need to prepare 927 mL of 1.20 M HCl?

What mass of sodium bicarbonate would be needed to neutralize the spill if a bottle containing 1.75 L of concentrated HCl dropped on a lab floor and broke open?

Answer 1

1.

Al(s) + AgNO3(aq) Ag(s) + Al(NO3)3(aq)

The balanced reaction equation is expressed as;

Al(s) + 3AgNO3(aq) 3Ag(s) + Al(NO3)3(aq)

Since this reaction takes place in water and since AgNO3 and Al(NO3)3 are both water soluble, the net ionic reaction is:

Al(s) + 3Ag⁺ (aq) 3Ag(s) + Al⁺³(aq)

It is easy to see that Al is oxidized (it is more active) and silver ion is reduced (it is less active). The half reactions are:

Al Al⁺³ + 3e^- (oxidation)
Ag⁺ + e^- Ag (reduction)

2.

36.0% HCl by mass

So, in 100 g of solution contain 36 g of HCl.

Density of 1.18 g/mL

Volume = Mass / Density
= 100 g / (1.18 g/mL)
= 84.75 mL

So, in 84.75 mL of solution, the mass of HCl = 36 g
or,  in 1 mL of solution, the mass of HCl = (36/84.75) g
or,  in 1000 mL of solution, the mass of HCl = 1000 x (36/84.75) g
= 424.78 g

Molar mass of HCl = 36.5 g/mol
So, 36.5 g of HCl = 1 mol
or, 1 g of HCl = (1/36.5) mol
or, 424.78 g of HCl = (424.78 / 36.5) mol = 11.64 mol

Since, 11.64 mol of HCl is present in 1000 mL (or 1 L) , so the molarity = 11.64 M

(b)

Let the volume of 11.64 M solution be V.

We know that
M1V1 = M2V2
or, 11.64 M x V = 1.20 M x 927 mL
or, V = (1.20 M x 927 mL) / 11.64 M
or, V = 95.57 mL

Take 95.57 mL of 11.64 M solution and mark it to 927 mL.

(c)

1.75 L of 11.64 M HCl

Since, Molarity = Moles / Liter

So, moles of HCl = 11.64 M x 1.75 L = 20.37 moles

Now, HCl and NaHCO3 reacts as the reaction shown below;

NaHCO3(aq) + HCl(aq) NaCl(aq) + CO2(g) + H2O(l)

In the above balanced reaction equation
1 moles of HCl reacts with 1 mole of NaHCO3.

So, moles of NaHCO3 required to neutralize the whole HCl = 20.37 moles

Molar mass of NaHCO3 = 84 g/mol
So, 1 mole of NaHCO3 = 84 g
or, 20.37 moles of NaHCO3 = 20.37 x 84 g = 1711 g