practice: balancing equations #1\nshow all work on a separate sheet of paper. place coefficient final…

practice: balancing equations #1\nshow all work on a separate sheet of paper. place coefficient final answers in the blanks provided. you can leave coefficients of 1 blank.\n1. 2 h₂ + ___ o₂ → ___ h₂o\n2. ___ h₃po₄ + 3 koh → ___ k₃po₄ + 3 h₂o\n3. 6 k + ___ b₂o₃ → 3 k₂o + 2 b\n4. ___ hcl + ___ naoh → ___ nacl + ___ h₂o\n5. 10 na + 2 nano₃ → 6 na₂o + ___ n₂\n6. 4 c + ___ s₈ → 4 cs₂\n7. 2 na + ___ o₂ → ___ na₂o₂\n8. 2 n₂ + 5 o₂ → 2 n₂o₅\n9. 2 h₃po₄ + 3 mg(oh)₂ → ___ mg₃(po₄)₂ + 6 h₂o\n10. 2 naoh + ___ h₂co₃ → 2 na₂co₃ + 2 h₂o\n11. ___ koh + ___ hbr → ___ kbr + ___ h₂o\n12. ___ h₂ + ___ o₂ → ___ h₂o₂\n13. 4 na + ___ o₂ → 2 na₂o\n14. 2 al(oh)₃ + 3 h₂co₃ → ___ al₂(co₃)₃ + 6 h₂o\n15. 16 al + 3 s₈ → 8 al₂s₃\n16. 6 cs + ___ n₂ → 2 cs₃n\n17. ___ mg + ___ cl₂ → ___ mgcl₂\n18. 10 rb + 2 rbno₃ → 6 rb₂o + ___ n₂\n19. 2 c₆h₆ + 15 o₂ → 12 co₂ + 6 h₂o\n20. ___ n₂ + ___ h₂ → ___ nh₃\nreactions unit

practice: balancing equations #1\nshow all work on a separate sheet of paper. place coefficient final answers in the blanks provided. you can leave coefficients of 1 blank.\n1. 2 h₂ + ___ o₂ → ___ h₂o\n2. ___ h₃po₄ + 3 koh → ___ k₃po₄ + 3 h₂o\n3. 6 k + ___ b₂o₃ → 3 k₂o + 2 b\n4. ___ hcl + ___ naoh → ___ nacl + ___ h₂o\n5. 10 na + 2 nano₃ → 6 na₂o + ___ n₂\n6. 4 c + ___ s₈ → 4 cs₂\n7. 2 na + ___ o₂ → ___ na₂o₂\n8. 2 n₂ + 5 o₂ → 2 n₂o₅\n9. 2 h₃po₄ + 3 mg(oh)₂ → ___ mg₃(po₄)₂ + 6 h₂o\n10. 2 naoh + ___ h₂co₃ → 2 na₂co₃ + 2 h₂o\n11. ___ koh + ___ hbr → ___ kbr + ___ h₂o\n12. ___ h₂ + ___ o₂ → ___ h₂o₂\n13. 4 na + ___ o₂ → 2 na₂o\n14. 2 al(oh)₃ + 3 h₂co₃ → ___ al₂(co₃)₃ + 6 h₂o\n15. 16 al + 3 s₈ → 8 al₂s₃\n16. 6 cs + ___ n₂ → 2 cs₃n\n17. ___ mg + ___ cl₂ → ___ mgcl₂\n18. 10 rb + 2 rbno₃ → 6 rb₂o + ___ n₂\n19. 2 c₆h₆ + 15 o₂ → 12 co₂ + 6 h₂o\n20. ___ n₂ + ___ h₂ → ___ nh₃\nreactions unit

Answer

Explanation:

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

To balance the oxygen atoms, since there are 2 oxygen atoms in $O_2$ and 1 in $H_2O$, we put a 2 in front of $H_2O$. Then to balance hydrogen, we put a 2 in front of $H_2$. So the balanced equation is $2H_2+1O_2\rightarrow2H_2O$.

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

For potassium, since there are 3 potassium atoms in $K_3PO_4$, we put a 3 in front of $KOH$. Then for hydrogen and oxygen balance, the equation becomes $1H_3PO_4 + 3KOH\rightarrow1K_3PO_4+3H_2O$.

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

For potassium, to get 2 potassium atoms in $K_2O$, we put a 6 in front of $K$. For boron, we put a 2 in front of $B$ and 3 in front of $K_2O$. So it is $6K + 1B_2O_3\rightarrow3K_2O+2B$.

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

The number of atoms of each element is already the same on both sides in a 1:1:1:1 ratio. So it is $1HCl + 1NaOH\rightarrow1NaCl+1H_2O$.

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

First, balance nitrogen. To get 2 nitrogen atoms in $N_2$, we need 2 moles of $NaNO_3$. Then for sodium and oxygen balance, we get $10Na+2NaNO_3\rightarrow6Na_2O + 1N_2$.

Step6: Balance $C + S_8\rightarrow CS_2$

To balance sulfur, since there are 8 sulfur atoms in $S_8$, we need 4 moles of $CS_2$. Then for carbon, we put a 4 in front of $C$. So it is $4C+1S_8\rightarrow4CS_2$.

Step7: Balance $Na + O_2\rightarrow Na_2O_2$

For sodium, we need 2 moles of $Na$ to balance with $Na_2O_2$. So it is $2Na + 1O_2\rightarrow1Na_2O_2$.

Step8: Balance $N_2 + O_2\rightarrow N_2O_5$

To balance oxygen, since there are 5 oxygen atoms in $N_2O_5$, we need 5/2 moles of $O_2$. To get whole - number coefficients, we multiply all by 2, getting $2N_2+5O_2\rightarrow2N_2O_5$.

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

For magnesium, we need 3 moles of $Mg(OH)_2$. For phosphate, we need 2 moles of $H_3PO_4$. Then for hydrogen and oxygen balance, we get $2H_3PO_4+3Mg(OH)_2\rightarrow1Mg_3(PO_4)_2 + 6H_2O$.

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

For sodium, we need 2 moles of $NaOH$. Then for hydrogen and oxygen balance, we get $2NaOH+1H_2CO_3\rightarrow1Na_2CO_3+2H_2O$.

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

The equation is already balanced in a 1:1:1:1 ratio. So it is $1KOH + 1HBr\rightarrow1KBr+1H_2O$.

Step12: Balance $H_2 + O_2\rightarrow H_2O_2$

The equation is balanced as $1H_2+1O_2\rightarrow1H_2O_2$.

Step13: Balance $Na + O_2\rightarrow Na_2O$

For sodium, we need 4 moles of $Na$ to balance with 2 moles of $Na_2O$. So it is $4Na+1O_2\rightarrow2Na_2O$.

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

For aluminum, we need 2 moles of $Al(OH)_3$. For carbonate, we need 3 moles of $H_2CO_3$. Then for hydrogen and oxygen balance, we get $2Al(OH)_3+3H_2CO_3\rightarrow1Al_2(CO_3)_3+6H_2O$.

Step15: Balance $Al + S_8\rightarrow Al_2S_3$

For sulfur, to balance 8 sulfur atoms in $S_8$, we need 8 moles of $Al_2S_3$. Then for aluminum, we need 16 moles of $Al$. So it is $16Al+3S_8\rightarrow8Al_2S_3$.

Step16: Balance $Cs + N_2\rightarrow Cs_3N$

For nitrogen, we need 2 moles of $Cs_3N$ to balance 2 nitrogen atoms in $N_2$. Then for cesium, we need 6 moles of $Cs$. So it is $6Cs+1N_2\rightarrow2Cs_3N$.

Step17: Balance $Mg + Cl_2\rightarrow MgCl_2$

The equation is already balanced as $1Mg + 1Cl_2\rightarrow1MgCl_2$.

Step18: Balance $Rb + RbNO_3\rightarrow Rb_2O + N_2$

Similar to the sodium - nitrate reaction, we get $10Rb+2RbNO_3\rightarrow6Rb_2O+1N_2$.

Step19: Balance $C_6H_6 + O_2\rightarrow CO_2 + H_2O$

First, for carbon, we need 6 moles of $CO_2$ for 6 carbon atoms in $C_6H_6$. For hydrogen, we need 3 moles of $H_2O$. Then for oxygen balance, we need 15/2 moles of $O_2$. Multiplying all by 2 gives $2C_6H_6+15O_2\rightarrow12CO_2 + 6H_2O$.

Step20: Balance $N_2 + H_2\rightarrow NH_3$

To balance nitrogen, we have 2 nitrogen atoms in $N_2$. To balance hydrogen, we need 3 moles of $H_2$ and 2 moles of $NH_3$. So it is $1N_2+3H_2\rightarrow2NH_3$.

Answer:

  1. $2H_2+1O_2\rightarrow2H_2O$
  2. $1H_3PO_4 + 3KOH\rightarrow1K_3PO_4+3H_2O$
  3. $6K + 1B_2O_3\rightarrow3K_2O+2B$
  4. $1HCl + 1NaOH\rightarrow1NaCl+1H_2O$
  5. $10Na+2NaNO_3\rightarrow6Na_2O + 1N_2$
  6. $4C+1S_8\rightarrow4CS_2$
  7. $2Na + 1O_2\rightarrow1Na_2O_2$
  8. $2N_2+5O_2\rightarrow2N_2O_5$
  9. $2H_3PO_4+3Mg(OH)_2\rightarrow1Mg_3(PO_4)_2 + 6H_2O$
  10. $2NaOH+1H_2CO_3\rightarrow1Na_2CO_3+2H_2O$
  11. $1KOH + 1HBr\rightarrow1KBr+1H_2O$
  12. $1H_2+1O_2\rightarrow1H_2O_2$
  13. $4Na+1O_2\rightarrow2Na_2O$
  14. $2Al(OH)_3+3H_2CO_3\rightarrow1Al_2(CO_3)_3+6H_2O$
  15. $16Al+3S_8\rightarrow8Al_2S_3$
  16. $6Cs+1N_2\rightarrow2Cs_3N$
  17. $1Mg + 1Cl_2\rightarrow1MgCl_2$
  18. $10Rb+2RbNO_3\rightarrow6Rb_2O+1N_2$
  19. $2C_6H_6+15O_2\rightarrow12CO_2 + 6H_2O$
  20. $1N_2+3H_2\rightarrow2NH_3$