Calculation of excess
In an excess calculation you will be given
the quantities of two reactants
We can tackle excess calculations in four steps
Copy the equation and under the reactants write the
number of moles
For each of the two reactants calculate how many moles
are present
Choose one of the reactants and using its value calculate
how many moles of the other are needed to react
Decide if you have more than this quantity of the second
reactant
Mg
+
2HCl
MgCl2
+
H2
8.51 g of magnesium powder was added to
100 cm3 of 2 mol l-1 hydrochloric acid.
Decide by calculation which of reactants is in excess
Copy the equation and under the reactants write the
number of moles
Mg
1 mole
+
2HCl
MgCl2
+
H2
2 moles
We only need to consider the two reactants for this calculation
For the two reactants calculate how many moles are present
Mg
Number of moles =
Number of moles =
HCl
8.51
24.3
mass
gram formula mass
= 0.35
Number of moles = conc x volume in litres
Number of moles = 2 x 100/1000
Number of moles = 0.2
Choose one of the reactants and using its value calculate
how many moles of the other are needed to react
Use the value of 0.35 moles of Mg
Mg
1 mole
0.35 moles
+
2HCl
MgCl2
+
H2
2 moles
0.7 moles
We need 0.7 moles for all of the Magnesium to react
Do you have more than this quantity of the second reactant ?
From the equation
We need 0.7 moles of HCl
From the calculation
We have 0.2 moles of HCl
We do not have enough HCl, we have too much Mg
Mg is in Excess
6.3 g of Magnesium reacts with 200 cm3 of 1 mol l-1 H2SO4.
Calculate
reactant which
is in4 excess
Mg
+ Hthe
MgSO
+ H2
2SO4
1 mole + 1 mole
0.26
MgMg
For
+ 0.26
No.
6.3/24.3
= Formula
0.26 Mass
No.of
of moles == Mass
/ Gram
No. of moles concentration
= 1 x 200/1000
= 0.2in litres
For HH2SO
x Volume
2SO
4 4 No. of moles =
Need 0.26
Havemoles
only 0.2
of Hmole
to excess
H
react
2SO
2SO4will all Mg
Magnesium
is4 of
in