balance the following chemical equations.
1. ____ fe + ____ h₂so₄ → ____ fe₂(so₄)₃ + ____ h₂
2. ____ c₂h₆ + ____ o₂ → ____ h₂o + ____ co₂
3. ____ koh + ____ h₃po₄ → ____ k₃po₄ + ____ h₂o
4. ____ sno₂ + ____ h₂ → ____ sn + ____ h₂o

Question

balance the following chemical equations.
1. 	____ fe + ____ h₂so₄ → ____ fe₂(so₄)₃ + ____ h₂
2. 	____ c₂h₆ + ____ o₂ → ____ h₂o + ____ co₂
3. 	____ koh + ____ h₃po₄ → ____ k₃po₄ + ____ h₂o
4. 	____ sno₂ + ____ h₂ → ____ sn + ____ h₂o

Accepted Answer

# Explanation:
## Step1: Balance Fe atoms
For the equation $Fe + H_2SO_4
ightarrow Fe_2(SO_4)_3 + H_2$, in $Fe_2(SO_4)_3$ there are 2 Fe atoms. So we put 2 in front of Fe: $2Fe + H_2SO_4
ightarrow Fe_2(SO_4)_3 + H_2$
## Step2: Balance $SO_4$ groups
In $Fe_2(SO_4)_3$ there are 3 $SO_4$ groups. So we put 3 in front of $H_2SO_4$: $2Fe + 3H_2SO_4
ightarrow Fe_2(SO_4)_3 + H_2$
## Step3: Balance H atoms
In $3H_2SO_4$ there are 6 H atoms. So we put 3 in front of $H_2$: $2Fe + 3H_2SO_4 = Fe_2(SO_4)_3 + 3H_2$

## Step4: Balance C atoms for $C_2H_6+O_2
ightarrow H_2O + CO_2$
In $C_2H_6$ there are 2 C atoms. In $CO_2$, to balance C, if we assume the coefficient of $C_2H_6$ is 1, for 2 C atoms,  we initially think of coefficient 2 for $CO_2$. But let's use a better approach. Let the coefficient of $C_2H_6$ be $x$, $O_2$ be $y$, $H_2O$ be $z$ and $CO_2$ be $w$. From C balance: $2x = w$, from H balance: $6x=2z$ (or $z = 3x$). From O balance: $2y=z + 2w$. Substitute $z = 3x$ and $w = 2x$ into $2y=z + 2w$, we get $2y=3x+4x=7x$. Let $x = 2$, then $w = 4$, $z=6$, $y = 7$. So $2C_2H_6+7O_2 = 6H_2O + 4CO_2$

## Step5: Balance K atoms for $KOH+H_3PO_4
ightarrow K_3PO_4 + H_2O$
In $K_3PO_4$ there are 3 K atoms. So we put 3 in front of KOH: $3KOH+H_3PO_4
ightarrow K_3PO_4 + H_2O$
## Step6: Balance H and O atoms
In $3KOH$ there are 3 H and 3 O (from OH). In $H_3PO_4$ there are 3 H. In $K_3PO_4$ there are 4 O (from $PO_4$). The total H on left is $3 + 3=6$, on right in $H_2O$, if we put 3 in front of $H_2O$ (since $H_2O$ has 2 H per molecule, $3\times2 = 6$ H). For O: left side $3$ (from $KOH$)+4 (from $H_3PO_4$) = 7, right side $4$ (from $K_3PO_4$)+3 (from $3H_2O$) = 7. So $3KOH + H_3PO_4=K_3PO_4+3H_2O$

## Step7: Balance O atoms for $SnO_2+H_2
ightarrow Sn + H_2O$
In $SnO_2$ there are 2 O atoms. In $H_2O$, to balance O, if the coefficient of $SnO_2$ is 1,  the coefficient of $H_2O$ is 2. Then from H balance (since in $H_2O$ there are 4 H atoms),  the coefficient of $H_2$ is 2. And Sn is already balanced. So $SnO_2+2H_2=Sn + 2H_2O$

# Answer:
1. $2Fe + 3H_2SO_4 = Fe_2(SO_4)_3 + 3H_2$
2. $2C_2H_6+7O_2 = 6H_2O + 4CO_2$
3. $3KOH + H_3PO_4=K_3PO_4+3H_2O$
4. $SnO_2+2H_2=Sn + 2H_2O$

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QUESTION IMAGE

Question

Explanation:

Step1: Balance Fe atoms

Step2: Balance \(SO_4\) groups

Step3: Balance H atoms

Step4: Balance C atoms for \(C_2H_6+O_2

Step5: Balance K atoms for \(KOH+H_3PO_4

Step6: Balance H and O atoms

Step7: Balance O atoms for \(SnO_2+H_2

Answer: