Calculate stoichiometric volumes of gases at r.t.p , volumes of solutions and concentrations of solutions expressed in g / dm3 and mol / dm3, including conversion between cm3 and dm3
Conversion between dm3 and cm3
1dm3 of solution or gas = 1000cm3
To convert dm3 to cm3, take the value x1000
To convert cm3 to dm3, take the value to be divided by 1000
To convert g/dm3 to mol/dm3, take the value in g/dm3 divide by Mr
To convert mol/dm3 to g/dm3, take the value in mol/dm3 X Mr
Example: 200cm3 = 0.2dm3 and 5000cm3 = 5dm3
Concentration of solution can be expressed in g/dm3 (how much mass of the solute in 1 dm3 of solvent)
or mol/dm3 (how much moles of solute in 1dm3 of solution)
- Both involve volume in dm3 but formula for gas and solution are not the same.
Mole of gas at rtp = volume of gas in dm3 divide by 24dm3
Mole of solution = volume of solution in dm3 X concentration (mol/dm3)
Example: given 18g of sodium chloride is dissolved in 50cm3 of water.
Find concentration of the salt solution in g/dm3 and mol/dm3
50cm3 of solution contains 18g of NaCl
1cm3 of solution contains (18/50)g of NaCl
Since 1dm3 = 1000cm3, 1dm3 of solution contains (18/50) X 1000 = 360g/dm3
To find Concentration in mol/dm3 = 360/ 58.5 = 6.15mol/dm3
Question 1: find the volume of carbon dioxide produced at rtp when 2.4g of carbon is heated in excess oxygen.
Balanced chemical equation: C + O2 → CO2
Mole of carbon = 2.4/ 12 = 0.2mol
Mole of carbon dioxide is also 0.2mol.
Volume of carbon dioxide gas = mole x 24dm3 = 0.2 x24 = 4.8dm3
Question 2 : find the volume of hydrogen gas produced at rtp when 20cm3 of 0.5mol/dm3 of hydrochloric acid is added to excess magnesium.
Balanced equation: 2HCl + Mg → MgCl2 + H2
Mole of HCl = 0.02 X 0.5 = 0.01 mol
2 mol of HCl produces 1 mol of H2, 0.01 mol of HCl produces 0.005mol of H2
Volume of hydrogen has = 0.005 X 24dm3 = 0.12dm3
Question 3: find the volume of 0.1mol/dm3 of sodium hydroxide solution that reacts with 30cm3 of 0.5mol/dm3 of dilute nitric acid.
Balanced chemical equation: NaOH + HNO3 → NaNO3 + H2O
Mole of nitric acid = 0.03 x 0.5 = 0.015mol
Mole of NaOH is also 0.015mol.
Volume of NaOH = mole/concentration (Mol/dm3) = 0.015/ 0.1 = 0.15dm3
Question 4: find the concentration in mol/dm3 of 50cm3 of dilute hydrochloric acid which reacts with potassium to produce 480cm3 of hydrogen gas.
Balanced chemical equation: 2HCl + 2K → 2KCl + H2
Mole of Hydrogen gas = 0.48 /24 = 0.02mol
Mole of HCl= 2 x 0.02 = 0.04 mol
Concentration of HCl = mole/ volume in dm3 = 0.04/0.05 = 0.8mol/dm3