use with textbook pages 148 to 151.
balancing chemical equations
refer to page 149 for detailed instructions on how to balance chemical equations.
balance each chemical equation by inserting the correct coefficients, or numbers.
1. ___ni₃ → ___n₂ + ___i₂
2. ___fe₂o₃ + ___hcl → ___fecl₃ + ___h₂o
3. ___zn + ___cu₃n₂ → ___cu + ___zn₃n₂
4. ___pbcl₂ + ___nai → ___pbi₂ + ___nacl
5. ___h₂s + ___al → ___h₂ + ___al₂s₃
6. ___as + ___o₂ → ___as₂o₅
7. ___al + ___i₂ → ___all₃
8. ___hgo → ___hg + ___o₂
9. ___ba + ___hoh → ___h₂ + ___ba(oh)₂
10. ___k + ___br₂ → ___kbr
11. ___sio₂ + ___hf → ___sif₄ + ___h₂o
12. ___s + ___o₂ → ___so₃
13. ___cl₂ + ___febr₃ → ___fecl₃ + ___br₂
14. ___h₂ + ___f₂ → ___hf
15. ___li + ___h₂o → ___lioh + ___h₂
16. ___cul₂ + ___fe → ___fel₂ + ___cu
17. ___bn + ___f₂ → ___bf₃ + ___n₂
18. ___fecl₃ + ___ca(oh)₂ → ___fe(oh)₃ + ___cacl₂

Question

Accepted Answer

# Explanation:
## Step1: Balance the first equation $ \text{NI}_3
ightarrow\text{N}_2 + \text{I}_2 $
Count the number of atoms on both sides. For nitrogen ($N$): in $ \text{NI}_3$ there is 1 $N$ atom, in $ \text{N}_2$ there are 2 $N$ atoms. For iodine ($I$): in $ \text{NI}_3$ there are 3 $I$ atoms, in $ \text{I}_2$ there are 2 $I$ atoms. The least - common multiple of 2 (for $N$) and 3 (for $I$) is 6. So, $2\text{NI}_3
ightarrow\text{N}_2 + 3\text{I}_2$

## Step2: Balance the second equation $ \text{Fe}_2\text{O}_3+\text{HCl}
ightarrow\text{FeCl}_3+\text{H}_2\text{O}$
For iron ($Fe$): 2 on the left (in $ \text{Fe}_2\text{O}_3$) and 1 on the right (in $ \text{FeCl}_3$). For oxygen ($O$): 3 on the left (in $ \text{Fe}_2\text{O}_3$) and 1 on the right (in $ \text{H}_2\text{O}$). For hydrogen ($H$): 1 on the left (in $ \text{HCl}$) and 2 on the right (in $ \text{H}_2\text{O}$). For chlorine ($Cl$): 1 on the left (in $ \text{HCl}$) and 3 on the right (in $ \text{FeCl}_3$).
First, balance $Fe$: $ \text{Fe}_2\text{O}_3 + 6\text{HCl}
ightarrow2\text{FeCl}_3+3\text{H}_2\text{O}$

## Step3: Balance the third equation $ \text{Zn}+\text{Cu}_3\text{N}_2
ightarrow\text{Cu}+\text{Zn}_3\text{N}_2$
For zinc ($Zn$): 1 on the left and 3 on the right. For copper ($Cu$): 3 on the left and 1 on the right. For nitrogen ($N$): 2 on the left and 2 on the right. So, $3\text{Zn}+\text{Cu}_3\text{N}_2
ightarrow3\text{Cu}+\text{Zn}_3\text{N}_2$

## Step4: Balance the fourth equation $ \text{PbCl}_2+\text{NaI}
ightarrow\text{PbI}_2+\text{NaCl}$
For lead ($Pb$): 1 on both sides. For chlorine ($Cl$): 2 on the left and 1 on the right. For sodium ($Na$): 1 on the left and 1 on the right. For iodine ($I$): 1 on the left and 2 on the right. So, $ \text{PbCl}_2 + 2\text{NaI}
ightarrow\text{PbI}_2+2\text{NaCl}$

## Step5: Balance the fifth equation $ \text{H}_2\text{S}+\text{Al}
ightarrow\text{H}_2+\text{Al}_2\text{S}_3$
For hydrogen ($H$): 2 on the left (in $ \text{H}_2\text{S}$) and 2 on the right (in $ \text{H}_2$). For sulfur ($S$): 1 on the left (in $ \text{H}_2\text{S}$) and 3 on the right (in $ \text{Al}_2\text{S}_3$). For aluminum ($Al$): 1 on the left and 2 on the right. So, $3\text{H}_2\text{S}+2\text{Al}
ightarrow3\text{H}_2+\text{Al}_2\text{S}_3$

## Step6: Balance the sixth equation $ \text{As}+\text{O}_2
ightarrow\text{As}_2\text{O}_5$
For arsenic ($As$): 1 on the left and 2 on the right. For oxygen ($O$): 2 on the left and 5 on the right. The least - common multiple of 2 (for $As$) and 5 (for $O$) is 10. So, $4\text{As}+5\text{O}_2
ightarrow2\text{As}_2\text{O}_5$

## Step7: Balance the seventh equation $ \text{Al}+\text{I}_2
ightarrow\text{AlI}_3$
For aluminum ($Al$): 1 on the left and 1 on the right. For iodine ($I$): 2 on the left and 3 on the right. The least - common multiple of 2 and 3 is 6. So, $2\text{Al}+3\text{I}_2
ightarrow2\text{AlI}_3$

## Step8: Balance the eighth equation $ \text{HgO}
ightarrow\text{Hg}+\text{O}_2$
For mercury ($Hg$): 1 on both sides. For oxygen ($O$): 1 on the left and 2 on the right. So, $2\text{HgO}
ightarrow2\text{Hg}+\text{O}_2$

## Step9: Balance the ninth equation $ \text{Ba}+\text{HOH}
ightarrow\text{H}_2+\text{Ba(OH)}_2$
For barium ($Ba$): 1 on the left and 1 on the right. For hydrogen ($H$): 2 on the left (in $ \text{HOH}$) and 2 on the right (in $ \text{H}_2$ and $ \text{Ba(OH)}_2$). For oxygen ($O$): 1 on the left (in $ \text{HOH}$) and 2 on the right (in $ \text{Ba(OH)}_2$). So, $ \text{Ba}+2\text{HOH}
ightarrow\text{H}_2+\text{Ba(OH)}_2$

## Step10: Balance the tenth equation $ \text{K}+\text{Br}_2
ightarrow\text{KBr}$
For potassium ($K$): 1 on the left and 1 on the right. For bromine ($Br$): 2 on the left and 1 on the right. So, $2\text{K}+\text{Br}_2
ightarrow2\text{KBr}$

## Step11: Balance the eleventh equation $ \text{SiO}_2+\text{HF}
ightarrow\text{SiF}_4+\text{H}_2\text{O}$
For silicon ($Si$): 1 on both sides. For oxygen ($O$): 2 on the left and 1 on the right. For hydrogen ($H$): 1 on the left and 2 on the right. For fluorine ($F$): 1 on the left and 4 on the right. So, $ \text{SiO}_2+4\text{HF}
ightarrow\text{SiF}_4+2\text{H}_2\text{O}$

## Step12: Balance the twelfth equation $ \text{S}+\text{O}_2
ightarrow\text{SO}_3$
For sulfur ($S$): 1 on both sides. For oxygen ($O$): 2 on the left and 3 on the right. The least - common multiple of 2 and 3 is 6. So, $2\text{S}+3\text{O}_2
ightarrow2\text{SO}_3$

## Step13: Balance the thirteenth equation $ \text{Cl}_2+\text{FeBr}_3
ightarrow\text{FeCl}_3+\text{Br}_2$
For iron ($Fe$): 1 on both sides. For chlorine ($Cl$): 2 on the left and 3 on the right. For bromine ($Br$): 3 on the left and 2 on the right. The least - common multiple of 2 (for $Cl$) and 3 (for $Br$) is 6. So, $3\text{Cl}_2 + 2\text{FeBr}_3
ightarrow2\text{FeCl}_3+3\text{Br}_2$

## Step14: Balance the fourteenth equation $ \text{H}_2+\text{F}_2
ightarrow\text{HF}$
For hydrogen ($H$): 2 on the left and 1 on the right. For fluorine ($F$): 2 on the left and 1 on the right. So, $ \text{H}_2+\text{F}_2
ightarrow2\text{HF}$

## Step15: Balance the fifteenth equation $ \text{Li}+\text{H}_2\text{O}
ightarrow\text{LiOH}+\text{H}_2$
For lithium ($Li$): 1 on the left and 1 on the right. For hydrogen ($H$): 2 on the left (in $ \text{H}_2\text{O}$) and 3 on the right (in $ \text{LiOH}$ and $ \text{H}_2$). For oxygen ($O$): 1 on the left (in $ \text{H}_2\text{O}$) and 1 on the right (in $ \text{LiOH}$). So, $2\text{Li}+2\text{H}_2\text{O}
ightarrow2\text{LiOH}+\text{H}_2$

## Step16: Balance the sixteenth equation $ \text{CuI}_2+\text{Fe}
ightarrow\text{FeI}_2+\text{Cu}$
For copper ($Cu$): 1 on both sides. For iron ($Fe$): 1 on both sides. For iodine ($I$): 2 on both sides. So, $ \text{CuI}_2+\text{Fe}
ightarrow\text{FeI}_2+\text{Cu}$

## Step17: Balance the seventeenth equation $ \text{BN}+\text{F}_2
ightarrow\text{BF}_3+\text{N}_2$
For boron ($B$): 1 on both sides. For nitrogen ($N$): 1 on the left and 2 on the right. For fluorine ($F$): 2 on the left and 3 on the right. The least - common multiple of 2 (for $N$) and 3 (for $F$) is 6. So, $2\text{BN}+3\text{F}_2
ightarrow2\text{BF}_3+\text{N}_2$

## Step18: Balance the eighteenth equation $ \text{FeCl}_3+\text{Ca(OH)}_2
ightarrow\text{Fe(OH)}_3+\text{CaCl}_2$
For iron ($Fe$): 1 on both sides. For chlorine ($Cl$): 3 on the left and 2 on the right. For calcium ($Ca$): 1 on the left and 1 on the right. For oxygen ($O$): 2 on the left (in $ \text{Ca(OH)}_2$) and 3 on the right (in $ \text{Fe(OH)}_3$). For hydrogen ($H$): 2 on the left (in $ \text{Ca(OH)}_2$) and 3 on the right (in $ \text{Fe(OH)}_3$). The least - common multiple of 2 (for $Cl$) and 3 (for $OH$) is 6. So, $2\text{FeCl}_3 + 3\text{Ca(OH)}_2
ightarrow2\text{Fe(OH)}_3+3\text{CaCl}_2$

# Answer:
1. $2\text{NI}_3
ightarrow\text{N}_2 + 3\text{I}_2$
2. $ \text{Fe}_2\text{O}_3 + 6\text{HCl}
ightarrow2\text{FeCl}_3+3\text{H}_2\text{O}$
3. $3\text{Zn}+\text{Cu}_3\text{N}_2
ightarrow3\text{Cu}+\text{Zn}_3\text{N}_2$
4. $ \text{PbCl}_2 + 2\text{NaI}
ightarrow\text{PbI}_2+2\text{NaCl}$
5. $3\text{H}_2\text{S}+2\text{Al}
ightarrow3\text{H}_2+\text{Al}_2\text{S}_3$
6. $4\text{As}+5\text{O}_2
ightarrow2\text{As}_2\text{O}_5$
7. $2\text{Al}+3\text{I}_2
ightarrow2\text{AlI}_3$
8. $2\text{HgO}
ightarrow2\text{Hg}+\text{O}_2$
9. $ \text{Ba}+2\text{HOH}
ightarrow\text{H}_2+\text{Ba(OH)}_2$
10. $2\text{K}+\text{Br}_2
ightarrow2\text{KBr}$
11. $ \text{SiO}_2+4\text{HF}
ightarrow\text{SiF}_4+2\text{H}_2\text{O}$
12. $2\text{S}+3\text{O}_2
ightarrow2\text{SO}_3$
13. $3\text{Cl}_2 + 2\text{FeBr}_3
ightarrow2\text{FeCl}_3+3\text{Br}_2$
14. $ \text{H}_2+\text{F}_2
ightarrow2\text{HF}$
15. $2\text{Li}+2\text{H}_2\text{O}
ightarrow2\text{LiOH}+\text{H}_2$
16. $ \text{CuI}_2+\text{Fe}
ightarrow\text{FeI}_2+\text{Cu}$
17. $2\text{BN}+3\text{F}_2
ightarrow2\text{BF}_3+\text{N}_2$
18. $2\text{FeCl}_3 + 3\text{Ca(OH)}_2
ightarrow2\text{Fe(OH)}_3+3\text{CaCl}_2$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step1: Balance the first equation \( \text{NI}_3

Step2: Balance the second equation \( \text{Fe}_2\text{O}_3+\text{HCl}

Step3: Balance the third equation \( \text{Zn}+\text{Cu}_3\text{N}_2

Step4: Balance the fourth equation \( \text{PbCl}_2+\text{NaI}

Step5: Balance the fifth equation \( \text{H}_2\text{S}+\text{Al}

Step6: Balance the sixth equation \( \text{As}+\text{O}_2

Step7: Balance the seventh equation \( \text{Al}+\text{I}_2

Step8: Balance the eighth equation \( \text{HgO}

Step9: Balance the ninth equation \( \text{Ba}+\text{HOH}

Answer: