Question

Copper reacts with hot and concentrated sulfuric acid to form copper sulfate, sulfur dioxide gas and water. Write f-number and balancing order of each element?

Answer 1

Hint: We all know that to balance a chemical equation we need to make sure that the number of atoms of each element on the reactant side should be equal to the number of atoms of each element on the product side. For this we need to make both sides equal, so we will multiply the number of atoms in each element until both sides are equal.

Complete step by step answer:
For balancing elements, it is good to neglect the pure elements until the end. So, we should first start balancing from carbon and hydrogen. Then, after that we need to balance oxygen. We need to know that it is easy to see that one mole of butane will produce four moles of carbon dioxide and five moles of water.
According to the question, it is given that copper reacts with hot and concentrated sulfuric acid to form copper sulfate, sulfur dioxide gas and water. We can write the equation involved in the reaction as,
$Cu + {H_2}S{O_4} \to CuS{O_4} + S{O_2} + {H_2}O$ (Unbalanced)
So, we will start balancing the above equation. For this we will first start balancing the elements on the left hand-side by putting 2 before sulphuric acid. Then, the equation becomes
$Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + {H_2}O$
Now, to balance the equation we will balance the elements on the right hand-side by putting 2 before the water molecule. We can write the balanced equation as,
$Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O$ (Balanced)

Note: To balance elements with charge we need to add electrons on one side of each half-reaction to balance the charge. For which we need to multiply the electrons by the two half-reactions to get the charge to balance out. It's fine to change coefficients as long as you change them on both sides of the equation.