Reaction Quantities
What would be the mass of ammonia produced in the reaction between
3000 g hydrogen with excess nitrogen?
The quantity of products is always determined by the reactant which is
NOT in excess.
N2 + 3H2
1. Balanced equation
2. Moles ratio
So…
3. Gram formula masses
And
therefore…
4. What 1 would give
Which
means…
But we
5. What you actually have have…
2NH3
3 moles
2 moles
3x2g
6g
2 x 17 g
34 g
1g
3000 g
34 g
6
34 g x 3000
6
= 1.7 kg
Calculating Excess
• 15g of calcium carbonate was reacted with 100 cm3 of 4 mol l-1
hydrochloric acid. A) Show which reactant is in excess.
CaCO3 + 2HCl
CaCl2 + CO2 + H2O
From equation, 1 mol CaCO3 requires 2 mol HCl for complete reaction
Moles CaCO3
100g
1g
15g
Moles HCl
1 mol
1/100 mol
15 x 1/100 mol
0.15 mol
Therefore 0.3 mol HCl (2 x 0.15) are
required for complete reaction
n = cv
n = 4 x 0.1
n = 0.4 mol
3 mol was required, 0.4 mol is
available, therefore HCl is in excess
(by 0.1 mol)
Continuing the question…
• 15g of calcium carbonate was reacted with 100 cm3 of 4 mol l-1
hydrochloric acid. B) Calculate the mass of carbon dioxide produced.
CaCO3 + 2HCl
Moles ratio
But actually have…
Convert into mass:
CaCl2 + CO2 + H2O
1
1
0.15
0.15
1 mol
0.15 mol
44g
6.6 g
Calculating Excess 2
• 0.6 g of magnesium ribbon was added to 40 cm3 2 mol l-1 hydrochloric
acid. A) Show which reactant is in excess.
Mg + 2 HCl
MgCl2 + H2
From equation, 1 mol Mg requires 2 mol HCl for complete reaction
Moles Mg
24.3g
1g
0.6g
Moles HCl
1 mol
1/24.3 mol
0.6 x 1/24.3 mol
0.025 mol
Therefore 0.05 mol HCl (2 x 0.025)
are required for complete reaction
n = cv
n = 2 x 0.04
n = 0.08 mol
0.05 mol was required, 0.08 mol is
available, therefore HCl is in excess
(by 0.03 mol)
Continuing the problem…
• 0.6 g of magnesium ribbon was added to 40 cm3 2 mol l-1 hydrochloric
acid. B) Calculate the mass of MgCl2 produced.
Mg + 2 HCl
Moles ratio
But actually have…
Convert into mass:
1
0.025
MgCl2 + H2
1
0.025
1 mol
0.025 mol
58.9g
1.47 g