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Question: How many grams of oxygen are produced by the decomposition of \(1\) mole of potassium chlorate \[KCl...

How many grams of oxygen are produced by the decomposition of 11 mole of potassium chlorate KClO3KCl{{O}_{3}} in the reaction 2KCl3→2KCl+3O22KC{{l}_{3}}\to 2KCl+3{{O}_{2}} ?

Explanation

Solution

We know that by arranging components and by expanding mass by determining measure of elements in grams. We can decide amounts of components in grams by changing over particles which incorporate atoms, molecules into grams. We need to change the molecule, atom, into grams with assistance of stoichiometry.

Complete step-by-step answer:
Here the given chemical reaction is 2KClO3→2KCl+3O22KCl{{O}_{3}}\to 2KCl+3{{O}_{2}}this equation is a balanced chemical reaction. Stoichiometry refers to evaluation of product and reactant taking part in any chemical reaction. Whereas stoichiometry is established on law of conservation for mass where the entire mass of reactant is of equivalent whole mass product which gives the relation among product and reactant usually makes the ratio a positive integer. Thus this shows that if you know the number of individual products you can easily calculate product amount.

Again if quantity of one reactant is known and product quantity can be determined using an experiment, then calculation of other reactants is also possible. This is represented in the following example of a balanced chemical reaction.

The reaction will be given by:
⇒5 gm of 2KClO3 × (1 mol of 2KClO312.55g of 2KClO3) × (3 mol of O22 mol of 2KClO3) × ( 16 g of O21 mol of O2)\Rightarrow 5\text{ }gm\text{ }of\text{ }2KCl{{O}_{3}}\text{ }\times \text{ }\left( \dfrac{1\text{ }mol\text{ }of\text{ }2KCl{{O}_{3}}}{12.55g\text{ }of\text{ }2KCl{{O}_{3}}} \right)\text{ }\times \text{ }\left( \dfrac{3\text{ }mol\text{ }of\text{ }{{O}_{2}}}{2\text{ }mol\text{ }of\text{ }2KCl{{O}_{3}}} \right)\text{ }\times \text{ }\left( \text{ }\dfrac{16\text{ }g\text{ }of\text{ }{{O}_{2}}}{1\text{ }mol\text{ }of\text{ }{{O}_{2}}} \right)

By solving the above equation by a simple division method we acquire grams of oxygen produced by decomposition of 5gram5gram of potassium chlorate.

⇒ 1.96 grams of O2\Rightarrow \text{ }1.96\text{ }grams\text{ }of\text{ }{{O}_{2}}

Note: Note that we need to recall that atomic mass in units is changed over to grams by multiplying Stoichiometry. We need to realize variation of atoms with distinctive weight. Along these lines one mole of atom in gram too. By chance that we don't change units over to required units there would be mistakes in conclusive answers.