10. c₃h₅oh + o₂ → co₂ + h₂o\n11. mg + hcl → mgcl₂ + h₂\n12. h₂so₄ + naoh → na₂so₄ + h₂o\n13. agno₃ + nacl →…

10. c₃h₅oh + o₂ → co₂ + h₂o\n11. mg + hcl → mgcl₂ + h₂\n12. h₂so₄ + naoh → na₂so₄ + h₂o\n13. agno₃ + nacl → agcl + nano₃\n14. bacl₂ + na₂so₄ → baso₄ + nacl\n15. p₄ + o₂ → p₂o₅\n16. k + h₂o → koh + h₂\n17. nh₃ + o₂ → no + h₂o\n18. h₃po₄ + ca(oh)₂ → ca₃(po₄)₂ + h₂o

10. c₃h₅oh + o₂ → co₂ + h₂o\n11. mg + hcl → mgcl₂ + h₂\n12. h₂so₄ + naoh → na₂so₄ + h₂o\n13. agno₃ + nacl → agcl + nano₃\n14. bacl₂ + na₂so₄ → baso₄ + nacl\n15. p₄ + o₂ → p₂o₅\n16. k + h₂o → koh + h₂\n17. nh₃ + o₂ → no + h₂o\n18. h₃po₄ + ca(oh)₂ → ca₃(po₄)₂ + h₂o

Answer

Explanation:

Step1: Balance equation 10

For the reaction $\mathrm{C}_2\mathrm{H}_5\mathrm{OH}+\mathrm{O}_2\rightarrow\mathrm{CO}_2+\mathrm{H}_2\mathrm{O}$, balance carbon first. There are 2 carbons in $\mathrm{C}_2\mathrm{H}_5\mathrm{OH}$, so put 2 in front of $\mathrm{CO}_2$. Then balance hydrogen. There are 6 hydrogens in $\mathrm{C}_2\mathrm{H}_5\mathrm{OH}$, so put 3 in front of $\mathrm{H}_2\mathrm{O}$. Finally, balance oxygen. There are 3 oxygen atoms in $\mathrm{C}_2\mathrm{H}_5\mathrm{OH}$, and after balancing carbon and hydrogen, the right - hand side has $2\times2 + 3\times1=7$ oxygen atoms. So, put 3 in front of $\mathrm{O}_2$. The balanced equation is $\mathrm{C}_2\mathrm{H}_5\mathrm{OH}+3\mathrm{O}_2\rightarrow2\mathrm{CO}_2 + 3\mathrm{H}_2\mathrm{O}$.

Step2: Balance equation 11

For $\mathrm{Mg}+\mathrm{HCl}\rightarrow\mathrm{MgCl}_2+\mathrm{H}_2$, there are 2 chlorines in $\mathrm{MgCl}_2$, so put 2 in front of $\mathrm{HCl}$. The balanced equation is $\mathrm{Mg}+2\mathrm{HCl}\rightarrow\mathrm{MgCl}_2+\mathrm{H}_2$.

Step3: Balance equation 12

For $\mathrm{H}_2\mathrm{SO}_4+\mathrm{NaOH}\rightarrow\mathrm{Na}_2\mathrm{SO}_4+\mathrm{H}_2\mathrm{O}$, there are 2 sodiums in $\mathrm{Na}_2\mathrm{SO}_4$, so put 2 in front of $\mathrm{NaOH}$. Then there are 4 hydrogens and 2 oxygens on the left - hand side from $2\mathrm{NaOH}$ and $\mathrm{H}_2\mathrm{SO}_4$, and 2 hydrogens and 1 oxygen in $\mathrm{H}_2\mathrm{O}$. The balanced equation is $\mathrm{H}_2\mathrm{SO}_4 + 2\mathrm{NaOH}\rightarrow\mathrm{Na}_2\mathrm{SO}_4+2\mathrm{H}_2\mathrm{O}$.

Step4: Balance equation 13

The equation $\mathrm{AgNO}_3+\mathrm{NaCl}\rightarrow\mathrm{AgCl}+\mathrm{NaNO}_3$ is already balanced as the number of each type of atom is the same on both sides.

Step5: Balance equation 14

For $\mathrm{BaCl}_2+\mathrm{Na}_2\mathrm{SO}_4\rightarrow\mathrm{BaSO}_4+\mathrm{NaCl}$, there are 2 chlorines in $\mathrm{BaCl}_2$ and 2 sodiums in $\mathrm{Na}_2\mathrm{SO}_4$, so put 2 in front of $\mathrm{NaCl}$. The balanced equation is $\mathrm{BaCl}_2+\mathrm{Na}_2\mathrm{SO}_4\rightarrow\mathrm{BaSO}_4 + 2\mathrm{NaCl}$.

Step6: Balance equation 15

For $\mathrm{P}_4+\mathrm{O}_2\rightarrow\mathrm{P}_2\mathrm{O}_5$, there are 4 phosphorous atoms in $\mathrm{P}_4$, so put 2 in front of $\mathrm{P}_2\mathrm{O}_5$. Then there are 10 oxygens in $2\mathrm{P}_2\mathrm{O}_5$, so put 5 in front of $\mathrm{O}_2$. The balanced equation is $\mathrm{P}_4+5\mathrm{O}_2\rightarrow2\mathrm{P}_2\mathrm{O}_5$.

Step7: Balance equation 16

For $\mathrm{K}+\mathrm{H}_2\mathrm{O}\rightarrow\mathrm{KOH}+\mathrm{H}_2$, to balance hydrogen, put 2 in front of $\mathrm{K}$, $\mathrm{H}_2\mathrm{O}$ and $\mathrm{KOH}$. The balanced equation is $2\mathrm{K}+2\mathrm{H}_2\mathrm{O}\rightarrow2\mathrm{KOH}+\mathrm{H}_2$.

Step8: Balance equation 17

For $\mathrm{NH}_3+\mathrm{O}_2\rightarrow\mathrm{NO}+\mathrm{H}_2\mathrm{O}$, first balance nitrogen (already 1 on each side). Then balance hydrogen. There are 3 hydrogens in $\mathrm{NH}_3$, so multiply $\mathrm{H}_2\mathrm{O}$ by $\frac{3}{2}$. Multiply all coefficients by 2 to get rid of the fraction. The balanced equation is $4\mathrm{NH}_3+5\mathrm{O}_2\rightarrow4\mathrm{NO}+6\mathrm{H}_2\mathrm{O}$.

Step9: Balance equation 18

For $\mathrm{H}_3\mathrm{PO}_4+\mathrm{Ca}(\mathrm{OH})_2\rightarrow\mathrm{Ca}_3(\mathrm{PO}_4)_2+\mathrm{H}_2\mathrm{O}$, there are 3 calciums in $\mathrm{Ca}_3(\mathrm{PO}_4)_2$, so put 3 in front of $\mathrm{Ca}(\mathrm{OH})_2$. There are 2 phosphates in $\mathrm{Ca}_3(\mathrm{PO}_4)_2$, so put 2 in front of $\mathrm{H}_3\mathrm{PO}_4$. Then balance hydrogen and oxygen. The balanced equation is $2\mathrm{H}_3\mathrm{PO}_4+3\mathrm{Ca}(\mathrm{OH})_2\rightarrow\mathrm{Ca}_3(\mathrm{PO}_4)_2 + 6\mathrm{H}_2\mathrm{O}$.

Answer:

  1. $\mathrm{C}_2\mathrm{H}_5\mathrm{OH}+3\mathrm{O}_2\rightarrow2\mathrm{CO}_2 + 3\mathrm{H}_2\mathrm{O}$
  2. $\mathrm{Mg}+2\mathrm{HCl}\rightarrow\mathrm{MgCl}_2+\mathrm{H}_2$
  3. $\mathrm{H}_2\mathrm{SO}_4 + 2\mathrm{NaOH}\rightarrow\mathrm{Na}_2\mathrm{SO}_4+2\mathrm{H}_2\mathrm{O}$
  4. $\mathrm{AgNO}_3+\mathrm{NaCl}\rightarrow\mathrm{AgCl}+\mathrm{NaNO}_3$
  5. $\mathrm{BaCl}_2+\mathrm{Na}_2\mathrm{SO}_4\rightarrow\mathrm{BaSO}_4 + 2\mathrm{NaCl}$
  6. $\mathrm{P}_4+5\mathrm{O}_2\rightarrow2\mathrm{P}_2\mathrm{O}_5$
  7. $2\mathrm{K}+2\mathrm{H}_2\mathrm{O}\rightarrow2\mathrm{KOH}+\mathrm{H}_2$
  8. $4\mathrm{NH}_3+5\mathrm{O}_2\rightarrow4\mathrm{NO}+6\mathrm{H}_2\mathrm{O}$
  9. $2\mathrm{H}_3\mathrm{PO}_4+3\mathrm{Ca}(\mathrm{OH})_2\rightarrow\mathrm{Ca}_3(\mathrm{PO}_4)_2 + 6\mathrm{H}_2\mathrm{O}$