balance the following chemical equations and show your work!
___sicl₄ + ___h₂o → ___ sio₂ + ___hcl
___as + ___ naoh→ ___ na₃aso₃ + ___ h₂
___zn + ___hcl → ___ zncl₂+ ___ h₂
___au₂s₃+ ___h₂ → ___au + ___h₂s
___n₂+ ___o₂ + ___h₂o→ ___hno₃

Question

Accepted Answer

# Explanation:
## Step1: Balance the first equation
- For $SiCl_{4}+H_{2}O
ightarrow SiO_{2}+HCl$
- Balance $Cl$ atoms: There are $4$ $Cl$ atoms in $SiCl_{4}$, so put $4$ in front of $HCl$. Now the equation is $SiCl_{4}+H_{2}O
ightarrow SiO_{2}+4HCl$
- Balance $H$ and $O$ atoms: There are $4$ $H$ atoms in $4HCl$, so put $2$ in front of $H_{2}O$. The balanced equation is $SiCl_{4}+2H_{2}O = SiO_{2}+4HCl$

## Step2: Balance the second equation
- For $As + NaOH
ightarrow Na_{3}AsO_{3}+H_{2}$
- Balance $Na$ atoms: There are $3$ $Na$ atoms in $Na_{3}AsO_{3}$, so put $3$ in front of $NaOH$. Now the equation is $As + 3NaOH
ightarrow Na_{3}AsO_{3}+H_{2}$
- Balance $H$ atoms: There are $3$ $H$ atoms in $3NaOH$, so put $\frac{3}{2}$ in front of $H_{2}$. To get rid of the fraction, multiply all coefficients by $2$. The balanced equation is $2As + 6NaOH = 2Na_{3}AsO_{3}+3H_{2}$

## Step3: Balance the third equation
- For $Zn + HCl
ightarrow ZnCl_{2}+H_{2}$
- Balance $Cl$ atoms: There are $2$ $Cl$ atoms in $ZnCl_{2}$, so put $2$ in front of $HCl$. The balanced equation is $Zn + 2HCl = ZnCl_{2}+H_{2}$

## Step4: Balance the fourth equation
- For $Au_{2}S_{3}+H_{2}
ightarrow Au + H_{2}S$
- Balance $S$ atoms: There are $3$ $S$ atoms in $Au_{2}S_{3}$, so put $3$ in front of $H_{2}S$. Now the equation is $Au_{2}S_{3}+H_{2}
ightarrow Au + 3H_{2}S$
- Balance $H$ atoms: There are $6$ $H$ atoms in $3H_{2}S$, so put $3$ in front of $H_{2}$. Balance $Au$ atoms: There are $2$ $Au$ atoms in $Au_{2}S_{3}$, so put $2$ in front of $Au$. The balanced equation is $Au_{2}S_{3}+3H_{2}=2Au + 3H_{2}S$

## Step5: Balance the fifth equation
- For $N_{2}+O_{2}+H_{2}O
ightarrow HNO_{3}$
- Balance $N$ atoms: There are $2$ $N$ atoms in $N_{2}$, so put $2$ in front of $HNO_{3}$. Now the equation is $N_{2}+O_{2}+H_{2}O
ightarrow 2HNO_{3}$
- Balance $H$ atoms: There are $2$ $H$ atoms in $2HNO_{3}$, so put $1$ in front of $H_{2}O$. Now the equation is $N_{2}+O_{2}+H_{2}O
ightarrow 2HNO_{3}$
- Balance $O$ atoms: There are $6$ $O$ atoms in $2HNO_{3}$, $2$ $O$ atoms in $O_{2}$ and $1$ $O$ atom in $H_{2}O$. So put $\frac{5}{2}$ in front of $O_{2}$. To get rid of the fraction, multiply all coefficients by $2$. The balanced equation is $2N_{2}+5O_{2}+2H_{2}O = 4HNO_{3}$

# Answer:
1. $1SiCl_{4}+2H_{2}O = 1SiO_{2}+4HCl$
2. $2As + 6NaOH = 2Na_{3}AsO_{3}+3H_{2}$
3. $Zn + 2HCl = ZnCl_{2}+H_{2}$
4. $Au_{2}S_{3}+3H_{2}=2Au + 3H_{2}S$
5. $2N_{2}+5O_{2}+2H_{2}O = 4HNO_{3}$

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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

Answer: