worked example (follow the steps and fill in the blanks)
equation: ( h_2 + o_2
ightarrow h_2o )
1. count atoms (initial): left — ( h = )____, ( o = )____. right — ( h = )____, ( o = )____
2. which element is unbalanced? __
3. add coefficient(s) to fix it: ( h_2 + o_2
ightarrow )____ ( h_2o )
4. recount and finish: left — ( h = )____, ( o = )____. right — ( h = )____, ( o = )____
5. final balanced equation: ____ ( h_2 + )____ ( o_2
ightarrow )____ ( h_2o )

guided practice (show work — write coefficients in front of each formula)
1. ____ ( h_2 + )____ ( cl_2
ightarrow )____ ( hcl )
2. ____ ( n_2 + )____ ( h_2
ightarrow )____ ( nh_3 )
3. ____ ( na + )____ ( cl_2
ightarrow )____ ( nacl )
4. ____ ( c_3h_8 + )____ ( o_2
ightarrow )____ ( co_2 + )____ ( h_2o )

Question

worked example (follow the steps and fill in the blanks)
equation: ( h_2 + o_2
ightarrow h_2o )
1. count atoms (initial): left — ( h = )____, ( o = )____. right — ( h = )____, ( o = )____
2. which element is unbalanced? __
3. add coefficient(s) to fix it: ( h_2 + o_2
ightarrow )____ ( h_2o )
4. recount and finish: left — ( h = )____, ( o = )____. right — ( h = )____, ( o = )____
5. final balanced equation: ____ ( h_2 + )____ ( o_2
ightarrow )____ ( h_2o )

guided practice (show work — write coefficients in front of each formula)
1. ____ ( h_2 + )____ ( cl_2
ightarrow )____ ( hcl )
2. ____ ( n_2 + )____ ( h_2
ightarrow )____ ( nh_3 )
3. ____ ( na + )____ ( cl_2
ightarrow )____ ( nacl )
4. ____ ( c_3h_8 + )____ ( o_2
ightarrow )____ ( co_2 + )____ ( h_2o )

Accepted Answer

# Explanation:
## Worked example
### Step1: Count initial atoms
For $H_2 + O_2
ightarrow H_2O$, on the left - hand side (LHS):
 - For hydrogen ($H$): In $H_2$, the number of $H$ atoms is $2$ (since the sub - script of $H$ in $H_2$ is $2$).
 - For oxygen ($O$): In $O_2$, the number of $O$ atoms is $2$ (sub - script of $O$ in $O_2$ is $2$).
On the right - hand side (RHS):
 - For hydrogen ($H$): In $H_2O$, the number of $H$ atoms is $2$ (sub - script of $H$ in $H_2O$ is $2$).
 - For oxygen ($O$): In $H_2O$, the number of $O$ atoms is $1$ (sub - script of $O$ in $H_2O$ is $1$).

### Step2: Identify unbalanced element
The oxygen atoms are unbalanced ($LHS:O = 2$, $RHS:O = 1$).

### Step3: Add coefficient to balance $O$
We add a coefficient of $2$ to $H_2O$ (so the equation becomes $H_2+O_2
ightarrow2H_2O$).

### Step4: Recount atoms
 - LHS: $H = 2$ (from $H_2$), $O = 2$ (from $O_2$).
 - RHS: In $2H_2O$, $H$ atoms $=2\times2 = 4$, $O$ atoms $=2\times1 = 2$.

### Step5: Add coefficient to balance $H$
We add a coefficient of $2$ to $H_2$ (the final balanced equation is $2H_2 + O_2
ightarrow2H_2O$).
 - LHS: $H=2\times2 = 4$, $O = 2$.
 - RHS: $H = 2\times2=4$, $O=2$.

## Guided practice
### 1. For $H_2+Cl_2
ightarrow HCl$
 - LHS: $H = 2$, $Cl = 2$.
 - RHS: $H = 1$, $Cl = 1$.
 - Add a coefficient of $2$ to $HCl$ (equation: $H_2+Cl_2
ightarrow2HCl$).
 - Check: LHS $H = 2$, $Cl = 2$; RHS $H=2\times1 = 2$, $Cl = 2\times1=2$.

### 2. For $N_2+H_2
ightarrow NH_3$
 - LHS: $N = 2$, $H = 2$.
 - RHS: $N = 1$, $H = 3$.
 - Add a coefficient of $2$ to $NH_3$ (equation becomes $N_2+H_2
ightarrow2NH_3$). Now RHS $N = 2$, $H=6$.
 - Add a coefficient of $3$ to $H_2$ (final equation $N_2 + 3H_2
ightarrow2NH_3$).
 - Check: LHS $N = 2$, $H=3\times2 = 6$; RHS $N = 2$, $H=2\times3 = 6$.

### 3. For $Na+Cl_2
ightarrow NaCl$
 - LHS: $Na = 1$, $Cl = 2$.
 - RHS: $Na = 1$, $Cl = 1$.
 - Add a coefficient of $2$ to $NaCl$ (equation $Na+Cl_2
ightarrow2NaCl$). Now RHS $Na = 2$, $Cl = 2$.
 - Add a coefficient of $2$ to $Na$ (final equation $2Na+Cl_2
ightarrow2NaCl$).
 - Check: LHS $Na = 2$, $Cl = 2$; RHS $Na = 2$, $Cl = 2$.

### 4. For $C_2H_6+O_2
ightarrow CO_2+H_2O$
 - Balance $C$ first: Add a coefficient of $2$ to $CO_2$ ($C_2H_6+O_2
ightarrow2CO_2+H_2O$).
 - Balance $H$: Add a coefficient of $3$ to $H_2O$ ($C_2H_6+O_2
ightarrow2CO_2 + 3H_2O$).
 - Now balance $O$:
   - RHS: In $2CO_2$, $O$ atoms $=2\times2 = 4$; in $3H_2O$, $O$ atoms $=3\times1 = 3$. Total $O$ on RHS $=4 + 3=7$.
   - LHS: Let the coefficient of $O_2$ be $\frac{7}{2}$ (but we want whole - number coefficients). Multiply the entire equation by $2$ (equation becomes $2C_2H_6+7O_2
ightarrow4CO_2+6H_2O$).
 - Check:
   - LHS: $C = 2\times2 = 4$, $H=2\times6 = 12$, $O=7\times2 = 14$.
   - RHS: $C = 4\times1 = 4$, $H=6\times2 = 12$, $O=(4\times2)+(6\times1)=8 + 6=14$.

# Answer:
## Worked example
1. Left — $H = 2$, $O = 2$; Right — $H = 2$, $O = 1$
2. Oxygen ($O$)
3. $2$
4. Left — $H = 2$, $O = 2$; Right — $H = 4$, $O = 2$
5. $2H_2+O_2
ightarrow2H_2O$

## Guided practice
1. $1H_2 + 1Cl_2
ightarrow2HCl$
2. $1N_2+3H_2
ightarrow2NH_3$
3. $2Na + 1Cl_2
ightarrow2NaCl$
4. $2C_2H_6+7O_2
ightarrow4CO_2+6H_2O$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Worked example

Step1: Count initial atoms

Step2: Identify unbalanced element

Step3: Add coefficient to balance \(O\)

Step4: Recount atoms

Step5: Add coefficient to balance \(H\)

Guided practice

1. For \(H_2+Cl_2

2. For \(N_2+H_2

3. For \(Na+Cl_2

4. For \(C_2H_6+O_2

Answer:

Worked example

Guided practice