practice calculations

Page 1

Practice

Calculations:

Write out the known and unknown data first.

Show all working and record answers to 3 significant figures.

1. An acid – base titration yielded the

following information.

1.25.0 mL of a 0.250 molL-1solution

of potassium hydroxide required 15.4 mL of sulfuric acid to be fully

neutralised. Using the equation below calculate the concentration of the

sulfuric acid in molL-1

2KOH + H2SO4 ® K2SO4 + 2H2O

2. An acid – base titration yielded the

following information.

13.7ml of 0.250molL-1 HCl was

required to fully neutralise 25ml of Ba(OH)2. Using the equation below calculate the

concentration of the Ba (OH)­2 in molL-1.

Ba(OH)2 + 2HCl ® BaCl2 + 2H2O

3.

Aspirin,

C9H8O4 , is made by reaction of:

Salicyclic Acid + Ethanoic anhydride à Aspirin

+ Ethanoic acid

C7­H6O3 + C4­H6O3

à C9H8O4 + C2­H4O2

How many grams of ethanoic

anhydride C4­H6O3 are needed to make one

aspirin tablet, which contains

0.33 g of aspirin?

4.

When

4.78g of an oxide of lead are heated in a stream of hydrogen, 4.14g of lead are

formed.

PbxOy +

H2 à

Pb + H2O

Determine the empirical formula of the lead oxide, PbxOy.

You need to work out the mass of the

lead and oxygen so that you can determine the empirical

formal of PbxOy

Help

The

mass of lead, Pb in PbxOy = 4.14g

The

mass of lead oxide, PbxOy = 4.78g

So

mass of oxygen in PbxOy must be 4.78g -

4.14g = 0.64 g

n(Ba(OH)2)

= 3.425 x 10‑3 mol = 1.7125 x 10-3

mol

c(Ba(OH)2) = = 0.0685molL-1

Q3)

M(C9H8O4)

= 180 g mol-1

(i) n(C9H8O4)

= m/M = 0.33 g / 180 g mol-1 = 1.83 x 10-3 mol

(ii) n(C4H6O3) =

n(C9H8O4) = 1:1 mole ratio so

n(C4H6O3)

1.83 x 10-3 mol

(iii) m(C4H6O3)

= 1.83 x 10-3 mol x 102 g mol-1 = 0.187 g

Q4)

Mass of lead oxide

is 4.78g

Mass of lead is

4.14g

So

mass of oxygen that is lost by the lead is worked out from

the

mass of the PbO – mass of Pb

The

mass of oxygen is 4.78g - 4.14g = 0.64 g

We

don’t need the mass of H2 or H2O

i.e. PbO

+ H2 à

Pb + H2O

(4.78g

+ 0.64 g) + H2 à 4.14g + H2O

i) n(Pb) = 4.14/207= 0.020 mol

(ii) m(O) = 4.78 - 4.14 = 0.64 g

(iii) n(O) = 0.64/16= 0.04 mol

(iv) Hence Pb : O = 1 : 2 and empirical formula is PbO2.

Answers:

Q1)

n(KOH) = 0.250 x 25/1000 = 0.00625 mol

n(H2SO4) = n(KOH)

n(H2 SO4) = x 0.00625 mol = 0.003125 mol

c(H2SO4) = = 0.203 molL-1

Q2)

n(HCl) = 0.250 x 0.0137 = 3.425 x 10‑3 mol

n(Ba(OH)2) = n(HCl)

The mass of oxygen is 4.78g - 4.14g = 0.64 g

We don’t need the mass of H2 or H2O

i.e. PbO + H2 à Pb + H2O

(4.78g + 0.64 g) + H2 à 4.14g + H2O

i) n(Pb) = 4.14/207= 0.020 mol

(ii) m(O) = 4.78 - 4.14 = 0.64 g

(iii) n(O) = 0.64/16= 0.04 mol

(iv) Hence Pb : O = 1 : 2 and empirical formula is PbO2.

practice calculations