Question

Balance the following reactions:
 $ SbC{{l}_{3}}+KI{{O}_{3}}+HCl\to SbC{{l}_{5}}+ICl+{{H}_{2}}O+KCl $
(A) $ 2SbC{{l}_{3}}+KI{{O}_{3}}+6HCl\to 2SbC{{l}_{5}}+ICl+3{{H}_{2}}O+KCl $
(B) $ SbC{{l}_{3}}+KI{{O}_{3}}+4HCl\to 2SbC{{l}_{5}}+ICl+3{{H}_{2}}O+KCl $
(C) $ 2SbC{{l}_{3}}+KI{{O}_{3}}+4HCl\to 2SbC{{l}_{5}}+5ICl+2{{H}_{2}}O+KCl $
(D) None of the above.

Answer 1

Hint: Follow four easy steps to balance a chemical equation- we write the unbalanced equation to show the reactants and products and then write the number of atoms of each element there are on each side of the reaction arrow. We add coefficients (the numbers in front of the formulas) so the number of atoms of each element is the same on both sides of the equation. It's easiest to balance the hydrogen and oxygen atoms last.

Complete step by step solution:
When atoms gain or lose electrons to yield ions, or combine with other atoms to form molecules, their symbols are modified or combined to generate chemical formulas that appropriately represent these species. Extending this symbolism to represent both the identities and the relative quantities of substances undergoing a chemical (or physical) change involves writing and balancing a chemical equation.
We follow four easy steps to balance a chemical equation:
We write down the unbalanced equation. The reactants always go on the left side of the arrow. A plus sign separates them. Next, there is an arrow indicating the direction of the reaction (reactants become products). The products are always on the right side of the arrow. The order in which the reactants and products are written is not important.
The equation given to us in the question is,
 $ SbC{{l}_{3}}+KI{{O}_{3}}+HCl\to SbC{{l}_{5}}+ICl+{{H}_{2}}O+KCl $ .
We write down how many atoms of each element there are on each side of the reaction arrow. To do this, keep in mind a subscript indicates the number of atoms. For example, $ C{{l}_{2}} $ has 2 atoms of chlorine. When there is no subscript, it means there is 1 atom. The equation is not balanced when the number of atoms on each side isn't the same!
In the next step, we add coefficients to balance mass in a chemical equation. When balancing equations, we never change subscripts. Coefficients are whole number multipliers. If, for example, we write $ 2{{H}_{2}}O $ , that means we have 2 times the number of atoms in each water molecule, which would be 4 hydrogen atoms and 2 oxygen atoms. There is a strategy that will help us balance equations more quickly. It is called balancing by inspection.
Basically, we look at how many atoms we have on each side of the equation and add coefficients to the molecules to balance out the number of atoms.
Balance atoms present in a single molecule of reactant and product first.
The oxidation number of $ Sb $ changes from 3 to 5. The change in the oxidation number of $ Sb $ is 2.
The oxidation number of $ I $ changes from 5 to 1. The change in the oxidation number of $ I $ is 4.
The increase in the oxidation number is balanced with decrease in the oxidation number by multiplying.
 $ 2SbC{{l}_{3}}+KI{{O}_{3}}+6HCl\to 2SbC{{l}_{5}}+ICl+{{H}_{2}}O+KCl $ .
Balance any oxygen or hydrogen atoms last.
O atoms are balanced by adding 2 water molecules on the product side.
 $ 2SbC{{l}_{3}}+KI{{O}_{3}}+6HCl\to 2SbC{{l}_{5}}+ICl+2{{H}_{2}}O+KCl $
Hydrogen atoms are balanced.
Thus, the balanced equation is, $ 2SbC{{l}_{3}}+KI{{O}_{3}}+6HCl\to 2SbC{{l}_{5}}+ICl+2{{H}_{2}}O+KCl $ .
Hence, Option A is the correct answer.

Note:
Conservation of Mass states mass isn't created or destroyed in a chemical reaction. We need to add coefficients in front of the chemical formulas to adjust the number of atoms so they will be the same on both sides.