Question

Help me....!!

✿ Chemistry
✿Question in attachment


☞Best Mod
☞Best Star

☞Best usser

ll CAN ANSWER ll

Thank you :)

​

Answer 1

Answer:

plz mark me as brainlist

Answer 2

$\huge{\underline{\underline{\mathsf{\red{Given\: Equations\::}}}}}$

• $\sf{K_2O\:+\:H_2O\:\dashrightarrow\:KOH}$

• $\sf{Al\:+\:O_2\:\dashrightarrow\:Al_2O_3}$

• $\sf{N_2\:+\:H_2\:\dashrightarrow\:NH_3}$

• $\sf{Fe\:+\:HCl\:\dashrightarrow\:FeCl_3\:+\:H_2}$

• $\sf{KClO_3\:\dashrightarrow\:KCl\:+\:O_2}$

$\huge{\underline{\underline{\mathsf{\blue{To\:do\::}}}}}}$

Equation balancing

$\huge{\underline{\underline{\mathsf{\pink{Answer\::}}}}}}$

This table shows the number of elements on LHS and RHS .

In balancing we have to equate both the sides .

$\begin{tabular}{|c|c|c|}\cline{1-3} Element & LHS & RHS \\\cline{1-3}K & 2 & 1 \\\cline{1-3}H & 2 & 1\\\cline{1-3}O & 2 & 1\\\cline{1-3}\end{tabular}$

In order to make the number of elements same on both the sides we multiply them with few constants ,

Let's understand this by doing this equation ,

According to the observation if we multiply 2 to the right hand side than the equation will be balanced .

• $\sf{K_2O\:+\:H_2O\:\dashrightarrow\:2KOH}$

As now the number of elements is equal we call it to be balanced ,

$\begin{tabular}{|c|c|c|}\cline{1-3} Element & LHS & RHS \\\cline{1-3}K & 2 & 2 \\\cline{1-3}H & 2 & 2\\\cline{1-3}O & 2 & 2\\\cline{1-3}\end{tabular}$

Thus , now the equation is balanced .

Now repeat the same procedure for the other parts ,

Part second ,

$\begin{tabular}{|c|c|c|}\cline{1-3} Element & LHS & RHS \\\cline{1-3}Al & 1 & 2 \\\cline{1-3}O & 2 & 3\\\cline{1-3}\end{tabular}$

Solution ,

• $\sf{4Al\:+\:3O_2\:\dashrightarrow\:2Al_2O_3}$

Part third ,

$\begin{tabular}{|c|c|c|}\cline{1-3} Element & LHS & RHS \\\cline{1-3}N & 2 & 1 \\\cline{1-3}H & 2 & 3\\\cline{1-3}\end{tabular}$

Solution ,

• $\sf{N_2\:+\:3H_2\:\dashrightarrow\:2NH_3}$

Part fourth ,

$\begin{tabular}{|c|c|c|}\cline{1-3} Element & LHS & RHS \\\cline{1-3}Fe & 1 & 1 \\\cline{1-3}Cl & 1 & 3\\\cline{1-3}H & 1 & 2\\\cline{1-3}\end{tabular}$

Solution ,

• $\sf{2Fe\:+\:6HCl\:\dashrightarrow\:2FeCl_3\:+\:3H_2}$

Part fifth ,

$\begin{tabular}{|c|c|c|}\cline{1-3} Element & LHS & RHS \\\cline{1-3}K & 1 & 1 \\\cline{1-3}Cl & 1 & 1\\\cline{1-3}O & 3 & 2\\\cline{1-3}\end{tabular}$

Solution ,

• $\sf{2KClO_3\:\dashrightarrow\:2KCl\:+\:3O_2}$

Note : Kindly use web to view the answer