unit 8 = balancing chemical equations worksheet\ndirections - balance the following equations by placing the…

unit 8 = balancing chemical equations worksheet\ndirections - balance the following equations by placing the correct coefficients in front of each element or compound using the boxes given. each coefficient (or box) = ½ point!\n1. □n₂ + □o₂ → □no\n2. □c + □cl₂ → □ccl₄\n3. □k₂(co₃) + □cacl₂ → □ca(co₃) + □kcl\n4. □i₂ + □k → □ki\n5. □bacl₂ + □na₂(so₄) → □nacl + □ba(so₄)\n6. □n₂ + □hf → □hf₃n₄\n7. □zn + □h(no₃) → □zn(no₃)₂ + □h₂\n8. □nabr + □cl₂ → □nacl + □br₂\n9. □pb(no₃)₂ + □nai → □pbi₂ + □nano₃\n10. □cl₂ + □nai → □i₂ + □nacl\n11. □fe(no₃)₃ + □kcl → □fecl₃ + □k(no₃)\n12. □na + □zn(so₄) → □zn + □na₂(so₄)\n13. □al₂o₃ + □hcl → □alcl₃ + □h₂o\n14. □bacl₂ + □(nh₄)₂co₃ → □(nh₄)cl + □ba(co₃)\n15. □ph₃ + □o₂ → □p₄o₁₀ + □h₂o\n16. □o₂ + □fe → □fe₃o₄\n17. □o₂ + □fe → □fe₂o₃\n18. □h₂ + □n₂ → □nh₃\n19. □kclo₃ → □kcl + □o₂\n20. □al + □cu(so₄) → □cu + □al₂(so₄)₃

unit 8 = balancing chemical equations worksheet\ndirections - balance the following equations by placing the correct coefficients in front of each element or compound using the boxes given. each coefficient (or box) = ½ point!\n1. □n₂ + □o₂ → □no\n2. □c + □cl₂ → □ccl₄\n3. □k₂(co₃) + □cacl₂ → □ca(co₃) + □kcl\n4. □i₂ + □k → □ki\n5. □bacl₂ + □na₂(so₄) → □nacl + □ba(so₄)\n6. □n₂ + □hf → □hf₃n₄\n7. □zn + □h(no₃) → □zn(no₃)₂ + □h₂\n8. □nabr + □cl₂ → □nacl + □br₂\n9. □pb(no₃)₂ + □nai → □pbi₂ + □nano₃\n10. □cl₂ + □nai → □i₂ + □nacl\n11. □fe(no₃)₃ + □kcl → □fecl₃ + □k(no₃)\n12. □na + □zn(so₄) → □zn + □na₂(so₄)\n13. □al₂o₃ + □hcl → □alcl₃ + □h₂o\n14. □bacl₂ + □(nh₄)₂co₃ → □(nh₄)cl + □ba(co₃)\n15. □ph₃ + □o₂ → □p₄o₁₀ + □h₂o\n16. □o₂ + □fe → □fe₃o₄\n17. □o₂ + □fe → □fe₂o₃\n18. □h₂ + □n₂ → □nh₃\n19. □kclo₃ → □kcl + □o₂\n20. □al + □cu(so₄) → □cu + □al₂(so₄)₃

Answer

Explanation:

Step1: Balance nitrogen and oxygen in $N_2 + O_2\rightarrow NO$

Set coefficients to make nitrogen and oxygen atoms equal on both sides. $N_2+O_2 = 2NO$

Step2: Balance carbon and chlorine in $C + Cl_2\rightarrow CCl_4$

We need 2 moles of $Cl_2$ to balance chlorine. So $C + 2Cl_2=CCl_4$

Step3: Balance $K_2(CO_3)+CaCl_2\rightarrow Ca(CO_3)+KCl$

For potassium balance, we need 2 moles of $KCl$. So $K_2(CO_3)+CaCl_2 = Ca(CO_3)+2KCl$

Step4: Balance $I_2 + K\rightarrow KI$

We need 2 moles of $K$ and $KI$. So $I_2+2K = 2KI$

Step5: Balance $BaCl_2+Na_2(SO_4)\rightarrow NaCl+Ba(SO_4)$

We need 2 moles of $NaCl$. So $BaCl_2+Na_2(SO_4)=2NaCl + Ba(SO_4)$

Step6: Balance $N_2+Hf\rightarrow Hf_3N_4$

We need 3 moles of $Hf$ and 2 moles of $N_2$. So $2N_2 + 3Hf=Hf_3N_4$

Step7: Balance $Zn+H(NO_3)\rightarrow Zn(NO_3)_2+H_2$

We need 2 moles of $H(NO_3)$. So $Zn + 2H(NO_3)=Zn(NO_3)_2+H_2$

Step8: Balance $NaBr+Cl_2\rightarrow NaCl+Br_2$

We need 2 moles of $NaBr$ and $NaCl$. So $2NaBr+Cl_2 = 2NaCl+Br_2$

Step9: Balance $Pb(NO_3)_2+NaI\rightarrow PbI_2+NaNO_3$

We need 2 moles of $NaI$ and $NaNO_3$. So $Pb(NO_3)_2+2NaI=PbI_2 + 2NaNO_3$

Step10: Balance $Cl_2+NaI\rightarrow I_2+NaCl$

We need 2 moles of $NaI$ and $NaCl$. So $Cl_2+2NaI = I_2+2NaCl$

Step11: Balance $Fe(NO_3)_3+KCl\rightarrow FeCl_3+K(NO_3)$

We need 3 moles of $KCl$ and $K(NO_3)$. So $Fe(NO_3)_3+3KCl=FeCl_3+3K(NO_3)$

Step12: Balance $Na+Zn(SO_4)\rightarrow Zn+Na_2(SO_4)$

We need 2 moles of $Na$. So $2Na+Zn(SO_4)=Zn + Na_2(SO_4)$

Step13: Balance $Al_2O_3+HCl\rightarrow AlCl_3+H_2O$

We need 6 moles of $HCl$, 2 moles of $AlCl_3$ and 3 moles of $H_2O$. So $Al_2O_3+6HCl = 2AlCl_3+3H_2O$

Step14: Balance $BaCl_2+(NH_4)_2CO_3\rightarrow (NH_4)Cl+Ba(CO_3)$

We need 2 moles of $(NH_4)Cl$. So $BaCl_2+(NH_4)_2CO_3=2(NH_4)Cl+Ba(CO_3)$

Step15: Balance $PH_3+O_2\rightarrow P_4O_{10}+H_2O$

We need 4 moles of $PH_3$, 8 moles of $O_2$, 1 mole of $P_4O_{10}$ and 6 moles of $H_2O$. So $4PH_3+8O_2=P_4O_{10}+6H_2O$

Step16: Balance $O_2+Fe\rightarrow Fe_3O_4$

We need 3 moles of $Fe$ and 2 moles of $O_2$. So $2O_2+3Fe=Fe_3O_4$

Step17: Balance $O_2+Fe\rightarrow Fe_2O_3$

We need 4 moles of $Fe$ and 3 moles of $O_2$. So $3O_2+4Fe = 2Fe_2O_3$

Step18: Balance $H_2+N_2\rightarrow NH_3$

We need 3 moles of $H_2$ and 2 moles of $NH_3$. So $3H_2+N_2=2NH_3$

Step19: Balance $KClO_3\rightarrow KCl+O_2$

We need 2 moles of $KClO_3$, 2 moles of $KCl$ and 3 moles of $O_2$. So $2KClO_3=2KCl+3O_2$

Step20: Balance $Al+Cu(SO_4)\rightarrow Cu+Al_2(SO_4)_3$

We need 2 moles of $Al$ and 3 moles of $Cu(SO_4)$. So $2Al+3Cu(SO_4)=3Cu+Al_2(SO_4)_3$

Answer:

  1. $1N_2+1O_2 = 2NO$
  2. $1C + 2Cl_2=1CCl_4$
  3. $1K_2(CO_3)+1CaCl_2 = 1Ca(CO_3)+2KCl$
  4. $1I_2+2K = 2KI$
  5. $1BaCl_2+1Na_2(SO_4)=2NaCl + 1Ba(SO_4)$
  6. $2N_2 + 3Hf=1Hf_3N_4$
  7. $1Zn + 2H(NO_3)=1Zn(NO_3)_2+1H_2$
  8. $2NaBr+1Cl_2 = 2NaCl+1Br_2$
  9. $1Pb(NO_3)_2+2NaI=1PbI_2 + 2NaNO_3$
  10. $1Cl_2+2NaI = 1I_2+2NaCl$
  11. $1Fe(NO_3)_3+3KCl=1FeCl_3+3K(NO_3)$
  12. $2Na+1Zn(SO_4)=1Zn + 1Na_2(SO_4)$
  13. $1Al_2O_3+6HCl = 2AlCl_3+3H_2O$
  14. $1BaCl_2+1(NH_4)_2CO_3=2(NH_4)Cl+1Ba(CO_3)$
  15. $4PH_3+8O_2=1P_4O_{10}+6H_2O$
  16. $2O_2+3Fe=1Fe_3O_4$
  17. $3O_2+4Fe = 2Fe_2O_3$
  18. $3H_2+1N_2=2NH_3$
  19. $2KClO_3=2KCl+3O_2$
  20. $2Al+3Cu(SO_4)=3Cu+1Al_2(SO_4)_3$