practice: balancing equation\nshow all work on a separate sheet of paper. place coefficient final answers…

practice: balancing equation\nshow all work on a separate sheet of paper. place coefficient final answers. leave coefficients of \1\ blank.\n1. h₂ + o₂ → h₂o\n2. h₃po₄ + koh → k₃po₄ + h₂o\n3. k + b₂o₃ → k₂o + b\n4. hcl + naoh → nacl + h₂o\n5. na + nano₃ → na₂o + n₂\n6. c + s₈ → cs₂\n7. na + o₂ → na₂o₂\n8. n₂ + o₂ → n₂o₅\n9. h₃po₄ + mg(oh)₂ → mg₃(po₄)₂\n10. naoh + h₂co₃ → na₂co₃\n11. koh + hbr → kbr + h₂o\n12. h₂ + o₂ → h₂o₂\n13. na + o₂ → na₂o\n14. al(oh)₃ + h₂co₃ → al₂(co₃)

practice: balancing equation\nshow all work on a separate sheet of paper. place coefficient final answers. leave coefficients of \1\ blank.\n1. h₂ + o₂ → h₂o\n2. h₃po₄ + koh → k₃po₄ + h₂o\n3. k + b₂o₃ → k₂o + b\n4. hcl + naoh → nacl + h₂o\n5. na + nano₃ → na₂o + n₂\n6. c + s₈ → cs₂\n7. na + o₂ → na₂o₂\n8. n₂ + o₂ → n₂o₅\n9. h₃po₄ + mg(oh)₂ → mg₃(po₄)₂\n10. naoh + h₂co₃ → na₂co₃\n11. koh + hbr → kbr + h₂o\n12. h₂ + o₂ → h₂o₂\n13. na + o₂ → na₂o\n14. al(oh)₃ + h₂co₃ → al₂(co₃)

Answer

Explanation:

Step1: Balance hydrogen in $H_2 + O_2\rightarrow H_2O$

We need 2 moles of $H_2$ to get 4 hydrogen - atoms on the left - hand side to match 2 moles of $H_2O$ on the right - hand side. So the balanced equation is $2H_2+O_2\rightarrow 2H_2O$.

Step2: Balance $H_3PO_4 + KOH\rightarrow K_3PO_4 + H_2O$

For potassium, we need 3 moles of $KOH$. For hydrogen and oxygen balance, we get $H_3PO_4 + 3KOH\rightarrow K_3PO_4+3H_2O$.

Step3: Balance $K + B_2O_3\rightarrow K_2O + B$

To balance potassium, we need 6 moles of $K$. Then the balanced equation is $6K + B_2O_3\rightarrow 3K_2O+2B$.

Step4: Balance $HCl + NaOH\rightarrow NaCl + H_2O$

The equation is already balanced as it is: $HCl + NaOH\rightarrow NaCl + H_2O$.

Step5: Balance $Na+NaNO_3\rightarrow Na_2O + N_2$

To balance nitrogen, we need 2 moles of $NaNO_3$. Then to balance sodium and oxygen, the balanced equation is $10Na + 2NaNO_3\rightarrow 6Na_2O+N_2$.

Step6: Balance $C + S_8\rightarrow CS_2$

We need 8 moles of $C$ and 4 moles of $CS_2$ to balance sulfur. The balanced equation is $8C + S_8\rightarrow 8CS_2$.

Step7: Balance $Na+O_2\rightarrow Na_2O_2$

The balanced equation is $2Na + O_2\rightarrow Na_2O_2$.

Step8: Balance $N_2 + O_2\rightarrow N_2O_5$

To balance oxygen, we need 5 moles of $O_2$ and 2 moles of $N_2O_5$. The balanced equation is $2N_2+5O_2\rightarrow 2N_2O_5$.

Step9: Balance $H_3PO_4+Mg(OH)_2\rightarrow Mg_3(PO_4)_2 + H_2O$

We need 2 moles of $H_3PO_4$ and 3 moles of $Mg(OH)_2$. Then the balanced equation is $2H_3PO_4 + 3Mg(OH)_2\rightarrow Mg_3(PO_4)_2+6H_2O$.

Step10: Balance $NaOH + H_2CO_3\rightarrow Na_2CO_3 + H_2O$

We need 2 moles of $NaOH$. The balanced equation is $2NaOH + H_2CO_3\rightarrow Na_2CO_3+2H_2O$.

Step11: Balance $KOH + HBr\rightarrow KBr + H_2O$

The equation is already balanced: $KOH + HBr\rightarrow KBr + H_2O$.

Step12: Balance $H_2+O_2\rightarrow H_2O_2$

The balanced equation is $H_2+O_2\rightarrow H_2O_2$.

Step13: Balance $Na + O_2\rightarrow Na_2O$

We need 4 moles of $Na$ and 2 moles of $Na_2O$. The balanced equation is $4Na+O_2\rightarrow 2Na_2O$.

Step14: Balance $Al(OH)_3 + H_2CO_3\rightarrow Al_2(CO_3)_3+H_2O$

We need 2 moles of $Al(OH)_3$ and 3 moles of $H_2CO_3$. Then the balanced equation is $2Al(OH)_3+3H_2CO_3\rightarrow Al_2(CO_3)_3 + 6H_2O$.

Answer:

  1. $2H_2+O_2\rightarrow 2H_2O$
  2. $H_3PO_4 + 3KOH\rightarrow K_3PO_4+3H_2O$
  3. $6K + B_2O_3\rightarrow 3K_2O+2B$
  4. $HCl + NaOH\rightarrow NaCl + H_2O$
  5. $10Na + 2NaNO_3\rightarrow 6Na_2O+N_2$
  6. $8C + S_8\rightarrow 8CS_2$
  7. $2Na + O_2\rightarrow Na_2O_2$
  8. $2N_2+5O_2\rightarrow 2N_2O_5$
  9. $2H_3PO_4 + 3Mg(OH)_2\rightarrow Mg_3(PO_4)_2+6H_2O$
  10. $2NaOH + H_2CO_3\rightarrow Na_2CO_3+2H_2O$
  11. $KOH + HBr\rightarrow KBr + H_2O$
  12. $H_2+O_2\rightarrow H_2O_2$
  13. $4Na+O_2\rightarrow 2Na_2O$
  14. $2Al(OH)_3+3H_2CO_3\rightarrow Al_2(CO_3)_3 + 6H_2O$