part d balance the following equations
1. __ca + __o₂ → __cao
2. __n₂ + __h₂ → __nh₃
3. __cu₂o + __c → __cu + __co₂
4. __h₂o₂ → __h₂o + __o₂
5. __na + __mgf₂ → __naf + __mg
6. __mg + __hcl → __mgcl₂ + __h₂
7. __cl₂ + __ki → __kcl + __i₂
8. __nacl → __na + __cl₂
9. __na + __o₂ → __na₂o
10. __na + __hcl → __h₂ + __nacl
11. __k + __cl₂ → __kcl
12 challenge: __c₂h₆ + __o₂ → __co₂ + __h₂o

Question

Accepted Answer

# Explanation:
## Step1: Balance the first equation
For $Ca + O_2
ightarrow CaO$, balance $O$ atoms first. Since there are $2$ $O$ atoms in $O_2$, we need $2$ $CaO$. Then balance $Ca$ atoms, so $2Ca+O_2
ightarrow 2CaO$

## Step2: Balance the second equation
For $N_2 + H_2
ightarrow NH_3$, balance $N$ atoms first. There are $2$ $N$ atoms in $N_2$, so we need $2$ $NH_3$. Then balance $H$ atoms. Since there are $6$ $H$ atoms in $2NH_3$, we need $3H_2$. So $N_2 + 3H_2
ightarrow 2NH_3$

## Step3: Balance the third equation
For $Cu_2O+C
ightarrow Cu + CO_2$, balance $O$ atoms. Multiply $Cu_2O$ by $2$ and $CO_2$ by $1$ to balance $O$. Then balance $Cu$ atoms. Since there are $4$ $Cu$ atoms in $2Cu_2O$, we need $4Cu$. So $2Cu_2O + C
ightarrow 4Cu+CO_2$

## Step4: Balance the fourth equation
For $H_2O_2
ightarrow H_2O+O_2$, balance $O$ atoms. Multiply $H_2O_2$ by $2$, $H_2O$ by $2$ and $O_2$ by $1$. So $2H_2O_2
ightarrow 2H_2O + O_2$

## Step5: Balance the fifth equation
For $Na+MgF_2
ightarrow NaF+Mg$, balance $F$ atoms. Multiply $NaF$ by $2$. Then balance $Na$ atoms. Multiply $Na$ by $2$. So $2Na+MgF_2
ightarrow 2NaF+Mg$

## Step6: Balance the sixth equation
For $Mg+HCl
ightarrow MgCl_2+H_2$, balance $Cl$ atoms. Multiply $HCl$ by $2$. Then balance $H$ atoms. Since there are $2$ $H$ atoms in $2HCl$, we need $1H_2$. So $Mg + 2HCl
ightarrow MgCl_2+H_2$

## Step7: Balance the seventh equation
For $Cl_2+KI
ightarrow KCl+I_2$, balance $Cl$ atoms. Multiply $KCl$ by $2$. Then balance $I$ atoms. Multiply $KI$ by $2$. So $Cl_2+2KI
ightarrow 2KCl+I_2$

## Step8: Balance the eighth equation
For $NaCl
ightarrow Na+Cl_2$, balance $Cl$ atoms. Multiply $NaCl$ by $2$. Then balance $Na$ atoms. Multiply $Na$ by $2$. So $2NaCl
ightarrow 2Na+Cl_2$

## Step9: Balance the ninth equation
For $Na+O_2
ightarrow Na_2O$, balance $O$ atoms. Multiply $Na_2O$ by $2$. Then balance $Na$ atoms. Multiply $Na$ by $4$. So $4Na+O_2
ightarrow 2Na_2O$

## Step10: Balance the tenth equation
For $Na+HCl
ightarrow H_2+NaCl$, balance $H$ atoms. Multiply $HCl$ by $2$ and $NaCl$ by $2$. Then balance $Na$ atoms. Multiply $Na$ by $2$. So $2Na+2HCl
ightarrow H_2+2NaCl$

## Step11: Balance the eleventh equation
For $K+Cl_2
ightarrow KCl$, balance $Cl$ atoms. Multiply $KCl$ by $2$. Then balance $K$ atoms. Multiply $K$ by $2$. So $2K+Cl_2
ightarrow 2KCl$

## Step12: Balance the twelfth equation
For $C_2H_6+O_2
ightarrow CO_2+H_2O$, balance $C$ atoms. Multiply $CO_2$ by $2$. Balance $H$ atoms. Multiply $H_2O$ by $3$. Then balance $O$ atoms. Multiply $O_2$ by $\frac{7}{2}$. To get rid of the fraction, multiply all coefficients by $2$. So $2C_2H_6+7O_2
ightarrow 4CO_2+6H_2O$

# Answer:
1. $2Ca+O_2
ightarrow 2CaO$
2. $N_2 + 3H_2
ightarrow 2NH_3$
3. $2Cu_2O + C
ightarrow 4Cu+CO_2$
4. $2H_2O_2
ightarrow 2H_2O + O_2$
5. $2Na+MgF_2
ightarrow 2NaF+Mg$
6. $Mg + 2HCl
ightarrow MgCl_2+H_2$
7. $Cl_2+2KI
ightarrow 2KCl+I_2$
8. $2NaCl
ightarrow 2Na+Cl_2$
9. $4Na+O_2
ightarrow 2Na_2O$
10. $2Na+2HCl
ightarrow H_2+2NaCl$
11. $2K+Cl_2
ightarrow 2KCl$
12. $2C_2H_6+7O_2
ightarrow 4CO_2+6H_2O$

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QUESTION IMAGE

Question

Explanation:

Step1: Balance the first equation

Step2: Balance the second equation

Step3: Balance the third equation

Step4: Balance the fourth equation

Step5: Balance the fifth equation

Step6: Balance the sixth equation

Step7: Balance the seventh equation

Step8: Balance the eighth equation

Step9: Balance the ninth equation

Step10: Balance the tenth equation

Step11: Balance the eleventh equation

Step12: Balance the twelfth equation

Answer: