Reaction Based Questions
We know that atoms combine in whole-number ratios to form compounds and also that reactants and products appear in chemical equations in simple ratios.
The relative proportions of elements in a compound or those of the reactants and products in a chemical reaction are known as stiochiometry.
We will see in the following sections how the percentage composition of a compound and also the masses and volumes taking part in a chemical reaction help us arrive the stiochiometry.
Problems Based on Chemical Equations
A chemical equation is a precise way of expressing a lot of information about the way a reaction takes place. The following steps will help you use this information to solve various problems.
1. Write a chemical equation to represent the reaction and balance it.
2. Write the number of moles of each reactant and product. These numbers are the same as the numbers of molecules or ions on both sides of the equati
3. Convert the number of moles of each reactant and product into mass or volume (if gaseous) as convenient or required.
4. Calculate the mass or volume of one or more of the reactants and products on the basis of the data given.
How well the method works can be easily seen in the following example.
Suppose you have to calculate the mass of water formed by the reaction between 11.2 L of H2 and 5.6 L of O2 taken at STP. Proceed as follows.
|2H2(g) +||O2 (g) →||2H2O(1)|
|2 mol||1 mol||2 mol|
|2 x 22.4L (STP)||22.4 L (STP)||2 x 18 g|
|22.4 L (STP)||11.2 L (STP)||18g|
|11.2 L (STP)||5.6 L (STP)||9 g|
So, the mass of the water formed in the reaction = 9 g.
Illustration 10: Hydrogen and chlorine react in the gram-atomic ratio of 1 : 1 to form HCl. If 4.0 g of hydrogen is available for the reaction, what is the mass of chlorine required for the complete reaction to take place (Ar of H = 1.0, Cl = 35.5)?
|Solution:||Ar of H = 1.0.||∴ the gram-atomic mass of H = 1.0 g.|
|Ar of Cl = 35.5.||∴ the gram-atomic mass of Cl = 35.0 g.|
The gram-atomic ratio for the reaction between hydrogen and chlorine
= 1 :1
⇒1 gram-atom of hydrogen requires 1 gram-atom of chlorine
⇒1.0g of hydrogen requires 35 g of chlorine
∴ 4.0 g of hydrogen requires 4.0 x 35.5 = 142.0g of chlorine.
Exercise 21: 2.5 gram-atoms of carbon are required for a particular reaction. How many grams of carbon would you take (At. mass of C = 12.0)?
Ans. 30 gms
Exercise 22: What is the volume of O2 required to burn 18 kg of C and of the CO2 product in the reaction? Assume that the volumes are measured at STP.
Ans. 33600 litres
Exercise 23: What is the volume of O2 (measured at STP) produced when 61.25 g of KClO3 is strongly heated (K = 39, Cl = 35.5, O = 16)?2KClO3 → 2KCl + 3O2
Ans. 16.8 litres
Exercise 24: How much KClO3 must be heated to produce as much O2 as required to burn 24 g of carbon
(K = 39, Cl = 35.5, O = 16)? 2KClO3 → 2KCl + 3O2 C + O2 → CO2
Ans. 163.33 gm
Exercise 25: MnO2 oxidises HCl to Cl2 and forms MnCl2. What is the number of moles of HCl and the mass of MnO2 required to produce 44.8 L of Cl2 at STP (Mn = 55, Cl = 35.5, O =16)?
MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O
Ans. 174 gm