DETERMINING A LIMITING REACTANT:
Consider following balanced equation:
Ca(OH)2(s) + 2 HCl(aq) CaCl2(aq) + 2H2O(l)
15 g of calcium hydroxide is combined with 75 cm3 of HCl solution with a concentration of
0.5 mol.dm-3.
With these type of questions, work to at least 2 significant figures or 3 decimal places.
Which of the reagents is the limiting reagent?
15 g Ca(OH)2
n = m/M
= 15 g / 74.092 g.mol-1
= 0.202 mol
75 cm3 of HCl with a concentration of 0.5 mol.dm-3
n = CV
= 0.5 mol.dm-3 x 0.075 dm3
= 3.75 x 10-2 mol
Checking for the limiting reactant:
We use the ratio of the reactants according to the balanced reaction.
2 𝑚𝑜𝑙 𝐻𝐶𝑙
0.202 mol Ca(OH)2 x 1 𝑚𝑜𝑙 𝐶𝑎(𝑂𝐻)2 = 0.404 mol HCl.
If all 0.202 mol of Ca(OH)2 was to react, we will need 0.404 mol HCl.
This is more than the number of moles that we currently have for HCl.
1 𝑚𝑜𝑙 𝐶𝑎(𝑂𝐻)2
3.75 x 10-2 mol HCl x = 1.88 x 10-2 mol Ca(OH)2.
2 𝑚𝑜𝑙 𝐻𝐶𝑙
If all 3.75 x 10 -2 mol of HCl was to react, we will need 1.88 x 10-2 mol of Ca(OH)2.
This is less than the number of moles that we have for Ca(OH)2.
Therefore, HCl is the limiting reactant.
Now, try out this example: Respiration is a chemical reaction according to the following
equation:
C6H12O6 + 6O2 6CO2 + 6H2O + energy
Given 25 grams of glucose and 40 grams of oxygen,
(a) Determine the limiting reactant.
(b) What mass of carbon dioxide will form?
Consider following balanced equation:
Ca(OH)2(s) + 2 HCl(aq) CaCl2(aq) + 2H2O(l)
15 g of calcium hydroxide is combined with 75 cm3 of HCl solution with a concentration of
0.5 mol.dm-3.
With these type of questions, work to at least 2 significant figures or 3 decimal places.
Which of the reagents is the limiting reagent?
15 g Ca(OH)2
n = m/M
= 15 g / 74.092 g.mol-1
= 0.202 mol
75 cm3 of HCl with a concentration of 0.5 mol.dm-3
n = CV
= 0.5 mol.dm-3 x 0.075 dm3
= 3.75 x 10-2 mol
Checking for the limiting reactant:
We use the ratio of the reactants according to the balanced reaction.
2 𝑚𝑜𝑙 𝐻𝐶𝑙
0.202 mol Ca(OH)2 x 1 𝑚𝑜𝑙 𝐶𝑎(𝑂𝐻)2 = 0.404 mol HCl.
If all 0.202 mol of Ca(OH)2 was to react, we will need 0.404 mol HCl.
This is more than the number of moles that we currently have for HCl.
1 𝑚𝑜𝑙 𝐶𝑎(𝑂𝐻)2
3.75 x 10-2 mol HCl x = 1.88 x 10-2 mol Ca(OH)2.
2 𝑚𝑜𝑙 𝐻𝐶𝑙
If all 3.75 x 10 -2 mol of HCl was to react, we will need 1.88 x 10-2 mol of Ca(OH)2.
This is less than the number of moles that we have for Ca(OH)2.
Therefore, HCl is the limiting reactant.
Now, try out this example: Respiration is a chemical reaction according to the following
equation:
C6H12O6 + 6O2 6CO2 + 6H2O + energy
Given 25 grams of glucose and 40 grams of oxygen,
(a) Determine the limiting reactant.
(b) What mass of carbon dioxide will form?