Question

part d - balancing practice
balance each equation:

1. h₂ + cl₂ → hcl
2. k + o₂ → k₂o
3. caco₃ → cao + co₂
4. zn + hcl → zncl₂ + h₂
5. c₂h₆ + o₂ → co₂ + h₂o

Explanation:

Step1: Balance hydrogen and chlorine in $H_2 + Cl_2

ightarrow HCl$
There are 2 hydrogen and 2 chlorine atoms on the left - hand side. So, we need 2 moles of $HCl$ on the right - hand side. The balanced equation is $H_2+Cl_2 = 2HCl$.

Step2: Balance potassium and oxygen in $K + O_2

ightarrow K_2O$
There are 2 oxygen atoms on the left - hand side and 1 on the right - hand side. Multiply $K_2O$ by 2 to get 2 oxygen atoms on the right. Then, we need 4 potassium atoms on the left. The balanced equation is $4K + O_2=2K_2O$.

Step3: Check $CaCO_3

ightarrow CaO + CO_2$
The number of calcium (Ca), carbon (C), and oxygen (O) atoms is the same on both sides. So, it is already balanced: $CaCO_3 = CaO+CO_2$.

Step4: Balance zinc, hydrogen, and chlorine in $Zn + HCl

ightarrow ZnCl_2 + H_2$
There are 2 chlorine atoms in $ZnCl_2$. So, we need 2 moles of $HCl$ on the left - hand side. The balanced equation is $Zn + 2HCl=ZnCl_2 + H_2$.

Step5: Balance carbon, hydrogen, and oxygen in $C_2H_6+O_2

ightarrow CO_2 + H_2O$
First, balance carbon: 2 moles of $CO_2$ for 2 carbon atoms in $C_2H_6$. Then balance hydrogen: 3 moles of $H_2O$ for 6 hydrogen atoms in $C_2H_6$. Now, count the oxygen atoms: on the right - hand side, we have 2×2 + 3×1=7 oxygen atoms. So, we need $\frac{7}{2}$ moles of $O_2$ on the left - hand side. Multiply through by 2 to get whole numbers. The balanced equation is $2C_2H_6 + 7O_2=4CO_2+6H_2O$.

Answer:

1. $H_2+Cl_2 = 2HCl$
2. $4K + O_2=2K_2O$
3. $CaCO_3 = CaO+CO_2$
4. $Zn + 2HCl=ZnCl_2 + H_2$
5. $2C_2H_6 + 7O_2=4CO_2+6H_2O$