Question

Answer All Questions.

1) All of the following are strong electrolytes EXCEPT :

a. Mg(HCO3)2

b. H2S

c. KNO2

d. Li2SO4

e. HClO4

2) Look at the following balanced equation:

2 C4H10 + 13 02 —> 8 CO2 + 10 H2O

If 18.5 grams of C4H10 reacts with 52.4 g O2, what is the theoretical yeild of CO2 in grams?

3) A 7.56 g sample of acetic acid, HC2H3O2, is dissolved in water. What is the molarity of Ba(OH)2 solution if by 130.0 mL or solution is needed to neutralize the acetic acid?

2 HC2H3O2 (aq) + Ba(OH)2 (aq) ——> 2 H2O (l) + Ba(C2H3O2) (aq)

4) What is the oxidation number of N in (NH4)2S?

Answer 1

1)

An electrolyte is a substance which can conduct electricity either due to the presence of free electrons(in case of metals) or due to the presence of free ions( acids, bases, salt solutions etc.)

Among the given options, all the compounds in option a, option c , option d, option e are ionic in nature. So in their aqueous solutions they all act as strong electrolytes.

Option b. H₂S is the weak electrolyte.

As it is a covalent compound and it is a weak acid ,it acts as a weak electrolyte.

Hence, answer is option b.

2)

Given balanced chemical equation is:

2 C4H10 + 13 02 —> 8 CO2 + 10 H2O

From this:

2 moles of butane C4H10 reacts with 13 moles of oxygen and gives 8 moles of CO₂.

that is:

2 x (58.12 g/mol) of butane reacts with 13 x (32 g/mol) of oxygen and gives 8 x (44 g./mol) of CO₂

=> 116.24 g/mol. of of butane reacts with 416 g/mol of oxygen and gives 352 g./mol of CO₂

First find the limiting reagent between the the reagents:

116.24 g. of of butane reacts with 416 g of oxygen from the balanced chemical equation

18.5 grams of C4H10 reacts with how many grams of O₂

=> 18.5 g. C4H10 X [416 g. oxygen / 116.24 g. C4H10 ]

=> 66.2g of oxygen is required.

But only 52.4 g O2 is present ,hence oxygen is the limiting reagent.

Second calculate the required amount of product using the limiting reagent:

416 g of oxygen gives 352 g. of CO₂

then 52.4 g O2 gives how much CO₂ ?

=> 52.4 g O2 X [ 352 g.CO₂ / 416 g. O2 ]

=> 44.3 g CO₂

Hence, 44.3 g CO₂ is produced.

3)

The balanced chemical equation for the reaction is:

2 HC2H3O2 (aq) + Ba(OH)2 (aq) ——> 2 H2O (l) + Ba(C2H3O2) (aq)

2 moles of acetic acid reacts with 1 mol. Ba(OH)₂

Let the amount of water in which 7.56g. of acetic acid dissolved be 1L.

Molarity of acetic acid (M1) = (its given mass / molecular mass)/ 1L

= (7.56g. / 60 g/mol ) /1L.

= 0.126M

Use the formula:

M₁ V₁ /n₁ = M₂ V₂ / n₂

0.126 M X 1000mL. / 2 mol = M₂ . 130mL / 1 mol

=> M₂ = 0.126 M X 1000mL. / 2 X 130mL

= 0.484M

4)

The oxidation number of N in (NH4)2S :

Let the oxidation number of N be x

H oxidation number = 1,

S oxidation number = -2

So,

2 x + 8 (+1) + (-2) = 0

=> 2x + 8 -2 =0

=> 2x = -6

=> x = -3

the oxidation number of N is -3.