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______h₃po₄ + ______koh → ______k₃po₄ + ______h₂o ______k + ______b₂o₃ …

Question

____h₃po₄ + __koh → __k₃po₄ + ____h₂o
____k + __b₂o₃ → __k₂o + ____b
____hcl + __naoh → __nacl + ____h₂o
____na + __nano₃ → __na₂o + ____n₂
____c + __s₈ → ____cs₂
____na + __o₂ → ____na₂o₂
____n₂ + __o₂ → ____n₂o₅
____h₃po₄ + __mg(oh)₂ → __mg₃(po₄)₂ + ____h₂o
____naoh + __h₂co₃ → __na₂co₃ + ____h₂o
____koh + __hbr → __kbr + ____h₂o
____h₂ + __o₂ → ____h₂o₂
____na + __o₂ → ____na₂o
____al(oh)₃ + __h₂co₃ → __al₂(co₃)₃ + ____h₂o

Explanation:

Step1: Balance the first equation ($\ce{H3PO4 + KOH -> K3PO4 + H2O}$)

  • Phosphorus (P): 1 on left, 1 on right (balanced).
  • Potassium (K): Let coefficient of $\ce{KOH}$ be $x$, coefficient of $\ce{K3PO4}$ be 1. So $x = 3$ (to balance K: $x = 3\times1$).
  • Hydrogen (H): From $\ce{H3PO4}$ (3 H) and $\ce{KOH}$ (3 H, since $x = 3$), total H on left: $3 + 3 = 6$. So coefficient of $\ce{H2O}$ is $3$ (since each $\ce{H2O}$ has 2 H, $6/2 = 3$).
  • Oxygen (O): Left: $4 + 3\times1 = 7$; Right: $4 + 3\times1 = 7$ (balanced).

So coefficients: $\ce{H3PO4}$: 1, $\ce{KOH}$: 3, $\ce{K3PO4}$: 1, $\ce{H2O}$: 3.

Step2: Balance the second equation ($\ce{K + B2O3 -> K2O + B}$)

  • Boron (B): 2 on left, so coefficient of $\ce{B}$ is 2.
  • Oxygen (O): 3 on left, so coefficient of $\ce{K2O}$ is 3 (each has 1 O, $3/1 = 3$).
  • Potassium (K): Coefficient of $\ce{K2O}$ is 3, so K on right: $3\times2 = 6$. Thus coefficient of $\ce{K}$ is 6.

Coefficients: $\ce{K}$: 6, $\ce{B2O3}$: 1, $\ce{K2O}$: 3, $\ce{B}$: 2.

Step3: Balance the third equation ($\ce{HCl + NaOH -> NaCl + H2O}$)

  • Hydrogen (H): 1 (from $\ce{HCl}$) + 1 (from $\ce{NaOH}$) = 2; on right, $\ce{H2O}$ has 2 H, so coefficient of $\ce{H2O}$ is 1.
  • Chlorine (Cl): 1 on left, 1 on right (balanced with $\ce{NaCl}$: 1).
  • Sodium (Na): 1 on left ($\ce{NaOH}$), 1 on right ($\ce{NaCl}$) (balanced).
  • Oxygen (O): 1 on left, 1 on right (balanced).

Coefficients: $\ce{HCl}$: 1, $\ce{NaOH}$: 1, $\ce{NaCl}$: 1, $\ce{H2O}$: 1.

Step4: Balance the fourth equation ($\ce{Na + NaNO3 -> Na2O + N2}$)

  • Nitrogen (N): 2 on right (from $\ce{N2}$), so coefficient of $\ce{NaNO3}$ is 2 (to balance N: 2 N on left).
  • Sodium (Na): From $\ce{Na}$ (let coefficient be $x$) and $\ce{NaNO3}$ (2 Na), total Na on left: $x + 2$. On right, $\ce{Na2O}$: let coefficient be $y$. So $x + 2 = 2y$.
  • Oxygen (O): From $\ce{NaNO3}$ (2×3 = 6 O), so $y = 6$ (since each $\ce{Na2O}$ has 1 O, $6/1 = 6$). Then $x + 2 = 2\times6 \implies x = 10$.
  • Check N: 2 (from 2 $\ce{NaNO3}$) → $\ce{N2}$: 1 (balanced, 2 N).

Coefficients: $\ce{Na}$: 10, $\ce{NaNO3}$: 2, $\ce{Na2O}$: 6, $\ce{N2}$: 1.

Step5: Balance the fifth equation ($\ce{C + S8 -> CS2}$)

  • Sulfur (S): 8 on left, so coefficient of $\ce{CS2}$ is 4 (since each $\ce{CS2}$ has 2 S, $8/2 = 4$).
  • Carbon (C): Coefficient of $\ce{C}$ is 4 (to balance C: 4 on left, 4 on right).

Coefficients: $\ce{C}$: 4, $\ce{S8}$: 1, $\ce{CS2}$: 4.

Step6: Balance the sixth equation ($\ce{Na + O2 -> Na2O2}$)

  • Sodium (Na): Let coefficient of $\ce{Na2O2}$ be 1, so coefficient of $\ce{Na}$ is 2 (2 Na on right).
  • Oxygen (O): 2 on left (from $\ce{O2}$), 2 on right (from $\ce{Na2O2}$) (balanced).

Coefficients: $\ce{Na}$: 2, $\ce{O2}$: 1, $\ce{Na2O2}$: 1.

Step7: Balance the seventh equation ($\ce{N2 + O2 -> N2O5}$)

  • Nitrogen (N): 2 on left, 2 on right (balanced).
  • Oxygen (O): 2 on left, 5 on right. Let coefficient of $\ce{O2}$ be $x$, coefficient of $\ce{N2O5}$ be 2 (to make O even on right: $2\times5 = 10$). Then $2x = 10 \implies x = 5$.

Coefficients: $\ce{N2}$: 2, $\ce{O2}$: 5, $\ce{N2O5}$: 2.

Step8: Balance the eighth equation ($\ce{H3PO4 + Mg(OH)2 -> Mg3(PO4)2 + H2O}$)

  • Phosphorus (P): 2 on right, so coefficient of $\ce{H3PO4}$ is 2.
  • Magnesium (Mg): 3 on right, so coefficient of $\ce{Mg(OH)2}$ is 3.
  • Hydrogen (H): From $\ce{H3PO4}$ (2×3 = 6 H) and $\ce{Mg(OH)2}$ (3×2 = 6 H), total H: 12. So coefficient of $\ce{H2O}$ is 6 (12/2 = 6).
  • Oxygen (O): Left: $2\times4 + 3\times2 = 8 + 6 = 14$; Right: $2\times4 + 6\times1 = 8 + 6 = 14$ (balanced).

Coefficients: $\ce{H3PO4}$: 2, $\ce{Mg(OH)2}$: 3, $\ce{…

Answer:

  1. $\boldsymbol{1}\ce{H3PO4 + 3KOH -> 1K3PO4 + 3H2O}$
  2. $\boldsymbol{6}\ce{K + 1B2O3 -> 3K2O + 2B}$
  3. $\boldsymbol{1}\ce{HCl + 1NaOH -> 1NaCl + 1H2O}$
  4. $\boldsymbol{10}\ce{Na + 2NaNO3 -> 6Na2O + 1N2}$
  5. $\boldsymbol{4}\ce{C + 1S8 -> 4CS2}$
  6. $\boldsymbol{2}\ce{Na + 1O2 -> 1Na2O2}$
  7. $\boldsymbol{2}\ce{N2 + 5O2 -> 2N2O5}$
  8. $\boldsymbol{2}\ce{H3PO4 + 3Mg(OH)2 -> 1Mg3(PO4)2 + 6H2O}$
  9. $\boldsymbol{2}\ce{NaOH + 1H2CO3 -> 1Na2CO3 + 2H2O}$
  10. $\boldsymbol{1}\ce{KOH + 1HBr -> 1KBr + 1H2O}$
  11. $\boldsymbol{1}\ce{H2 + 1O2 -> 1H2O2}$
  12. $\boldsymbol{4}\ce{Na + 1O2 -> 2Na2O}$
  13. $\boldsymbol{2}\ce{Al(OH)3 + 3H2CO3 -> 1Al2(CO3)3 + 6H2O}$