Question

Complete the following chemical equations:

$${\left( i \right){\text{ }}C{a_3}{P_2} + {H_2}O \to } \\\ {\left( {ii} \right){\text{ }}Cu + {H_2}S{O_4}\left( {conc.} \right) \to }$$

Answer 1

A chemical reaction refers to a phenomenon which leads to the chemical transition of one type of chemical substance to another. Chemical reactions involve changes with reference to the electrons positioning in the formation as well as decomposition of chemical bonds that exist between the atoms, while no change occurs to the nuclei i.e. no alteration to the elements present. The substances (one or more than one) initially participating in a chemical reaction are known as reactants. These chemical reactions undergo a chemical change, resulting in the formation of either one or more products that usually possess properties distinct from the reactants.

Complete Step by step answer: There are five basic types of chemical reactions which are combination, single-displacement, double-displacement, decomposition and combustion. By analysing the reactants as well as products of the given reactions will allow you to place them into one of these different types of reactions. Few reactions can fit even into more than one category. Now, let us look at the given reactions one by one:

$(i){\text{ }}C{a_3}{P_2} + {H_2}O \to$

Calcium phosphide ($C{a_3}{P_2}$) are the red-brown crystals or grey granular lumps. It generally reacts with water (hydrolyses) to form calcium hydroxide as well as phosphine ($P{H_3}$) (it is a flammable poisonous gas). Thus, the complete reaction becomes:
$(i){\text{ }}C{a_3}{P_2} + {H_2}O \to 3Ca{\left( {OH} \right)_2}{\text{ + }}2PH$
It is a double-displacement reaction (i.e. when two different reactants exchange their ions or radicals to form two entirely different compounds).
The second given reaction is $(ii){\text{ }}Cu + {H_2}S{O_4}\left( {conc.} \right) \to$
In this reaction, $Cu$ acts as a reducing agent while conc. ${H_2}S{O_4}$ behaves like an oxidising agent. Thus, $Cu$ does not displace the hydrogen from ${H_2}S{O_4}$ rather oxidised by the ${H_2}S{O_4}$. Now, this is not an acid-base reaction, but it is a redox reaction, as you see $Cu$ gets oxidised to $C{u^{2 + }}$ and ${H_2}S{O_4}$ gets reduced to $S{O_2}$. Thus, the complete reaction is written below:

$(ii)Cu + 2{H_2}S{O_4}\left( {conc.} \right) \to 2{H_2}O + CuS{O_4} + S{O_2}$

Note: A balanced chemical equation simply obeys the law of conservation of mass. Balancing the chemical equations is a significant guiding principle in chemistry. A balanced chemical equation helps you to predict the amount of reactants required and the amount of products formed.