balancing equations
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
5.
nh₃ +
o₂ →
no +
h₂o
6.
kno₃ +
h₂co₃ →
k₂co₃ +
hno₃
7.
b₂br₆ +
hno₃ →
b(no₃)₃ +
hbr

Question

Accepted Answer

# Explanation:
## Step1: Balance the first equation
For the equation $Fe + H_{2}SO_{4}\to Fe_{2}(SO_{4})_{3}+H_{2}$, balance $Fe$ atoms first. There are 2 $Fe$ atoms in $Fe_{2}(SO_{4})_{3}$, so put 2 in front of $Fe$. Then balance $SO_{4}$ groups. There are 3 $SO_{4}$ groups in $Fe_{2}(SO_{4})_{3}$, so put 3 in front of $H_{2}SO_{4}$. Finally, balance $H$ atoms. There are 6 $H$ atoms in $3H_{2}SO_{4}$, so put 3 in front of $H_{2}$.
## Step2: Balance the second equation
For the equation $C_{2}H_{6}+O_{2}\to H_{2}O + CO_{2}$, balance $C$ atoms first. There are 2 $C$ atoms in $C_{2}H_{6}$, so put 2 in front of $CO_{2}$. Balance $H$ atoms. There are 6 $H$ atoms in $C_{2}H_{6}$, so put 3 in front of $H_{2}O$. Now count $O$ atoms on the right - hand side ($3\times1 + 2\times2=7$). So put $\frac{7}{2}$ in front of $O_{2}$. To get rid of the fraction, multiply all coefficients by 2. So we have $2C_{2}H_{6}+7O_{2}\to6H_{2}O + 4CO_{2}$.
## Step3: Balance the third equation
For the equation $KOH+H_{3}PO_{4}\to K_{3}PO_{4}+H_{2}O$, balance $K$ atoms. There are 3 $K$ atoms in $K_{3}PO_{4}$, so put 3 in front of $KOH$. Then balance $H$ atoms. There are $3 + 3=6$ $H$ atoms on the left - hand side (from $3KOH$ and $H_{3}PO_{4}$), so put 3 in front of $H_{2}O$.
## Step4: Balance the fourth equation
For the equation $SnO_{2}+H_{2}\to Sn + H_{2}O$, balance $O$ atoms. There are 2 $O$ atoms in $SnO_{2}$, so put 2 in front of $H_{2}O$. Then balance $H$ atoms. There are 4 $H$ atoms in $2H_{2}O$, so put 2 in front of $H_{2}$.
## Step5: Balance the fifth equation
For the equation $NH_{3}+O_{2}\to NO + H_{2}O$, balance $H$ atoms first. There are 3 $H$ atoms in $NH_{3}$, so put $\frac{3}{2}$ in front of $H_{2}O$. Balance $N$ atoms. There is 1 $N$ atom in $NH_{3}$ and 1 in $NO$. Now count $O$ atoms on the right - hand side ($1+\frac{3}{2}=\frac{5}{2}$). So put $\frac{5}{4}$ in front of $O_{2}$. To get rid of fractions, multiply all coefficients by 4. So we have $4NH_{3}+5O_{2}\to4NO + 6H_{2}O$.
## Step6: Balance the sixth equation
For the equation $KNO_{3}+H_{2}CO_{3}\to K_{2}CO_{3}+HNO_{3}$, balance $K$ atoms. There are 2 $K$ atoms in $K_{2}CO_{3}$, so put 2 in front of $KNO_{3}$. Then balance $NO_{3}$ groups. There are 2 $NO_{3}$ groups in $2KNO_{3}$, so put 2 in front of $HNO_{3}$.
## Step7: Balance the seventh equation
For the equation $B_{2}Br_{6}+HNO_{3}\to B(NO_{3})_{3}+HBr$, balance $B$ atoms first. There are 2 $B$ atoms in $B_{2}Br_{6}$, so put 2 in front of $B(NO_{3})_{3}$. Then balance $NO_{3}$ groups. There are 6 $NO_{3}$ groups in $2B(NO_{3})_{3}$, so put 6 in front of $HNO_{3}$. Finally, balance $Br$ atoms. There are 6 $Br$ atoms in $B_{2}Br_{6}$, so put 6 in front of $HBr$.

# Answer:
1. $2Fe + 3H_{2}SO_{4}\to Fe_{2}(SO_{4})_{3}+3H_{2}$
2. $2C_{2}H_{6}+7O_{2}\to6H_{2}O + 4CO_{2}$
3. $3KOH + H_{3}PO_{4}\to K_{3}PO_{4}+3H_{2}O$
4. $SnO_{2}+2H_{2}\to Sn + 2H_{2}O$
5. $4NH_{3}+5O_{2}\to4NO + 6H_{2}O$
6. $2KNO_{3}+H_{2}CO_{3}\to K_{2}CO_{3}+2HNO_{3}$
7. $B_{2}Br_{6}+6HNO_{3}\to2B(NO_{3})_{3}+6HBr$

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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

Answer: