Get ready...

Are you all familiar with the stuff? Any queries? doubts? No isn't it. if so just think of the best replay for these question.

what are the products formed if any, at the anode and the cathode, when CuSO4(aq) is being electrolyzed using copper electrodes? Give half-equations.

state the products formed when concentrated aq.CaCl2 is being electrolyzed using inert electrodes. what will happen to the amount of the product formed at the anode if you use an active electrode such as iron in the anode?

How to work out a limiting reagent question..

Limiting reagent calculations are not much more difficult than any other calculations, there is just one step more - comparison of amounts of reacting substances.

Once we have balanced reaction equation we know what amounts of substances will react. Let's try the following reaction of calcium hydroxide synthesis:

CaO + H2O -> Ca(OH)2

One mole of calcium oxide reacts with one mole of water to produce one mole of calcium hydroxide. What will happen if we have one mole of calcium oxide and two moles of water? Well, nothing will change - still only one mole of CaO can react with H2O, so after the reaction there will be one mole of water left. We call the reagent that reacts to the end limiting reagent.

Let's try to find out how much calcium hydroxide will be produced when 10.00 g of calcium oxide reacts with 3.00 g of water.

First of all we have to check which one is limiting reagent. How much water will react with 10.0 grams of CaO?
CaO + H2O -> Ca(OH)2

56.08 g : 18.01 g

10.00 g : x g


56.08 g : 18.01 g = 10.0 g : x g

x = 10.00 g/56.08 g×18.01 g = 3.213 g

We have not enough water - 3.00 g only, when CaO needs 3.213 g, so water is a limiting reagent. How much calcium hydroxide will be produced from 3.00 g of water?


CaO + H2O -> Ca(OH)2

18.01 g : 74.09 g

3.00 g : x g


18.01 g : 74.09 g = 3.00 g : x g

x = 3.00 g/18.01 g×74.09 g = 12.34 g

That's the final answer.

Limiting reagent: answers

answer 1.(a)
NH4NO3=80, Na3PO4=164, (NH4)3PO4=149 [These are relative molecular mass]

NH4NO3 Na3PO4
3 : 1

(3x80)g : 164g
30g : x
x= 20.5g this value is lesser than 50g of Na3PO4 given in the question.it means NH4NO3 is in excess and Na3PO4 is the limiting reagent.

try out the answers for (b) and (c)

Molarity Exercise

For the questions on this worksheet, consider the following equation:

Ca(OH)2(s) + 2 HCl(aq)  CaCl2(aq) + 2 H2O(l)


1) What type of chemical reaction is taking place? _____________________

2) How many liters of 0.100 M HCl would be required to react completely with 5.00 grams of calcium hydroxide?








3) If I combined 15.0 grams of calcium hydroxide with 75.0 mL of 0.500 M HCl, how many grams of calcium chloride would be formed?













4) What is the limiting reagent from the reaction in problem #3? __________

5) How many grams of the excess reagent will be left over after the reaction in problem 3 is complete?

[note 1000cm3 = 1 Liter

Limiting Reagent Worksheet

For the following reactions, find the following:

a) Which of the reagents is the limiting reagent?
b) What is the maximum amount of each product that can be formed?
c) How much of the other reagent is left over after the reaction is complete?

1) Consider the following reaction:

3 NH4NO3 + Na3PO4  (NH4)3PO4 + 3 NaNO3

Answer the questions above, assuming we started with 30 grams of ammonium nitrate and 50 grams of sodium phosphate.

2) Consider the following reaction:

3 CaCO3 + 2 FePO4  Ca3(PO4)2 + Fe2(CO3)3

Answer the questions at the top of this sheet, assuming we start with 100 grams of calcium carbonate and 45 grams of iron (III) phosphate.

Test for nitrate ion

test for carbonate ion

Redox Titration: observe the colour change..