0.35g 100cm3
0.1mol/dm3
20cm3
0.15mol/dm3
From acid-base mole ratio:
Va = MbVbna
Manb
Va = 0.15mol/dm3 x 20cm3 x 1
0.1mol/dm3 x 1
Va = 30cm3
Therefore, the acid reacted with RCO3 can be obtained as follows:
100cm3 - 30cm3 = 70cm3
Number of moles of acid reacted0.1mol →1000cm3 ? mol → 70cm3 = 0.007mol of HClFrom the equation (i)
1 mole of RCO3 → 2 moles of HCl
? moles of RCO3 → 0.007 moles of HCl
This gives:
= 0.0035 moles of RCO3
Number of moles =mass of the substance
molar mass
Therefore,
Molar mass of RCO3 = mass of RCO3
number of moles of RCO3
= 0.35g
0.0035mol
= 100g/mol
To identify the element R
RCO3 = 100g/mol
(R + 12 + 16 X 3)g/mol = 100g/mol
R + 60 = 100
R = 100 - 60
R = 40
The relative atomic mass of the element R is 40.
The element R therefore is calcium
Ref:Volumetric analysis and Mole calculations
Question 2. A gaseous hydrocarbon X contains 20% hydrogen by mass. 7.5g of X occupy 5.6dm3 at S.T.P.Suggested answers
| Elements | C | H |
|---|---|---|
| Percentage composition | 80 | 20 |
| Relative atomic mass | 12 | 1 |
| % composition divide by R.A.M | 80/12 = 6.67 | 20/1 = 20 |
| Divide by smallest number | 6.67/6.67= 1 | 20/6.67 =3 |
| Whole number ratios | 1 | 3 |
Case 1: In terms of volume of a gas; Number of moles = volume of a gas
molar volume
= 5.6dm3
22.4dm3/mol
= 0.25mol
Case 2: In terms of mass of a gas;
Number of moles = mass of the gas
molar mass
The molar mass of the gas is not known
Therefore;
Number of moles = 7.5g
xg/mol
Because both are the ways of expressing the number of moles of the same gas:
therefore,
7.5g = 0.25mol
xg/mol
=7.5/0.25
=30
Molecular formula =(Empirical formula)n
where:
relative molecular mass of empirical formula ( CH3 )= 12 + (1x3) = 15
the value of n can be obtained by using this relation:
n = Mr of the compound
Mr of the empirical formula
n = 30
15
n = 2
To get molecular formula we insert the value of n to (CH3)2
The molecular formula is C3H6
iii. The name of this compound is propene.
Ref: empirical formula and molecular formula, Mole calculations and Organic chemistry
Question 3. 20cm3 of sodium hydroxide solution contain 8.0g/dm3 were required for complete neutralization 0f 0.18g of a dibasic acid. Calculate the relative molecular mass of the acid.
Data given
Volume of the NaOH solution = 20cm3
Concentration of NaOH soln = 8.0g/dm3
Mass of dibasic acid reacted = 0.18g
Relative molecular mass of dibasic acid = ?
Solution
Molar mass of NaOH = 23 + 16 + 1
= 40g/mol
Molarity of NaOH = concentration (g/dm3)
molar mass
= 8.0g/dm3
40g/mol
= 0.2mol/dm3
But only 20cm3 of the solution reacted
Number of moles of NaOH reacted = ?
from molarity = number of moles
volume (dm3)
number of moles = molarity x volume (dm3)
= 0.2mol/dm3 x 0.02dm3
= 0.004mol
Balanced chemical equation for the reaction:
2NaOH + H2X → Na2X + 2H2O
From the equation:
2mol of NaOH → 1mol of H2X
0.004mol of NaOH → ? mol of H2X
= 0.004mol x 1mol
2mol
= 0.002mol
These are moles of dibasic acid required to react with NaOH
from number of moles = mass
molar mass
therefore:
molar mass = mass
number of moles
= 0.18g
0.002mol
= 90g/mol
Therefore, the relative molecular mass of dibasic acid is 90.
Question 4. A doctor prescribed antacid tablets made from magnesium hydroxide to a student who was suffering from excess stomach acid:
The doctor prescribed antacid tablets to neutralize excess acid and relieve the student's discomfort caused by excess acid. This works because stomach acid is hydrochloric acid and magnesium hydroxide is base, when the student takes the antacid tablets, the magnesium hydroxide reacts with the excess hydrochloric acid in neutralization reaction and thereby relieving the suffering.
Mg(OH)2 + 2HCl → MgCl2 + 2H2O
Calculate the moles of excess HCl
Patient's acid concentration = 210.00millimolar = 0.2100mol/dm3 Normal acid concentration = 160.00millimolar = 0.1600mol/dm3 Volume of gastric juice = 0.50L Excess acid concentration = 0.2100mol/dm3 - 0.1600mol/dm3 = 0.05mol/dm3 (ii)Number of moles of excess HCl acid = molarity x volume = 0.05mol/dm3 x 0.5dm3 = 0.025mol The number of moles of Mg(OH)2 required to neutralize this acid: Mg(OH)2 + 2HCl → MgCl2 + 2H2O From the equation: 1mol of Mg(OH)2 → 2mol of HCl ? mol of Mg(OH)2 → 0.025mol of HCl number of moles of Mg(OH)2 = 0.0125mol (iii) The mass of Mg(OH)2 molar mass of Mg(OH)2 = 58g/mol from number of moles= mass of the substance molar mass mass = number of moles x molar mass = 0.0125mol x 58g/mol = 0.725g
1 tablet = 0.145g ? tablets = 0.725g = 5 tablets
(1 tablet = 12hours → for 1 day =24hours = 2tablets)
2 tablets = 1 day
5 tables = ? days?
2.5daysRef:Volumetric analysis and Mole calculations
Question 5. An acid solution was prepared by dissolving 767cm3 of hydrogen chloride gas in water at STP to make 250cm3 of solution. Only 18cm3 of this solution were enough to neutralize 25cm3 containing 9.864g of the compound YHCO3 per litre of solution. Find:
Suggested solution
Number of moles of HCl = Volume of the gas
molar volume at STP
= 767cm3
22400cm3/mol
= 0.034mol
Molarity of HCl = number of moles
volume of soln(dm3)
= 0.034mol
0.25dm3
= 0.136mol/dm3
Therefore molarity of acid, Ma is 0.136mol/dm3
Let us calculate the molarity of base:
Data given
Volume of acid (Va) = 18cm3
volume of base (Vb) = 25cm3
Balanced chemical equation:
HCl(aq) + YHCO3(aq) → YCl(aq) + H2O(l) + CO2(g)
na = 1 nb =1
From acid-base mole ratio:
MaVa = na
MbVb = nb
Mb =MaVanb
Vbna
= 0.136mol/dm3 x 18cm3 x 1
25cm3 x 1
= 0.0979mol/dm3
From molarity = concentration(g/dm3)
molar mass
Molar mass = concentration(g/dm3)
molarity
= 9.864g/dm3
0.0979mol/dm3
= 100.76g/mol
YHCO3 = 100.76g/mol
Y + 1 + 12 + (16 x 3)g/mol = 100.76g/mol
Y + 61 = 100.8
Y = 100.8 - 61
Y = 39.8
(ii) The metal Y might be potassium.