Question

1. hcn + cuso₄ → h₂so₄ + cu(cn)₂
2. gaf₃ + cs → csf + ga
3. bas + ptf₂ → baf₂ + pts
4. n₂ + h₂ → __ nh₃
5. naf + br₂ → nabr + f₂
6. pb(oh)₂ + hcl → h₂o + pbcl₂
7. albr₃ + k₂so₄ → kbr + al₂(so₄)₃
8. ch₄ + o₂ → co₂ + h₂o
9. na₃po₄ + cacl₂ → nacl + ca₃(po₄)₂
10. k + cl₂ → __ kcl
11. al + hcl → h₂ + alcl₃
12. n₂ + f₂ → __ nf₃
13. so₂ + li₂se → sse₂ + li₂o
14. nh₃ + h₂so₄ → __ (nh₄)₂so₄

Answer:

Let's solve these chemical equations by balancing them one by one. We'll use the method of balancing atoms on both sides of the equation.

Problem 13: $\boldsymbol{AlBr_3 + K_2SO_4

ightarrow KBr + Al_2(SO_4)_3}$

Step 1: Balance Al atoms

There are 2 Al atoms on the right (in $Al_2(SO_4)_3$), so we put a coefficient of 2 in front of $AlBr_3$ on the left.
$2AlBr_3 + K_2SO_4
ightarrow KBr + Al_2(SO_4)_3$

Step 2: Balance $SO_4^{2-}$ ions

There are 3 $SO_4^{2-}$ ions on the right (in $Al_2(SO_4)_3$), so we put a coefficient of 3 in front of $K_2SO_4$ on the left.
$2AlBr_3 + 3K_2SO_4
ightarrow KBr + Al_2(SO_4)_3$

Step 3: Balance K and Br atoms

Now, on the left, we have 6 K atoms (from $3K_2SO_4$) and 6 Br atoms (from $2AlBr_3$). So we put a coefficient of 6 in front of $KBr$ on the right.
$2AlBr_3 + 3K_2SO_4
ightarrow 6KBr + Al_2(SO_4)_3$

Problem 14: $\boldsymbol{CH_4 + O_2

ightarrow CO_2 + H_2O}$

Step 1: Balance C atoms

There is 1 C atom on both sides, so C is balanced for now.

Step 2: Balance H atoms

There are 4 H atoms in $CH_4$, so we put a coefficient of 2 in front of $H_2O$ to get 4 H atoms on the right.
$CH_4 + O_2
ightarrow CO_2 + 2H_2O$

Step 3: Balance O atoms

On the right, we have 2 (from $CO_2$) + 2 (from $2H_2O$) = 4 O atoms. So we put a coefficient of 2 in front of $O_2$ on the left.
$CH_4 + 2O_2
ightarrow CO_2 + 2H_2O$

Problem 15: $\boldsymbol{Na_3PO_4 + CaCl_2

ightarrow NaCl + Ca_3(PO_4)_2}$

Step 1: Balance $PO_4^{3-}$ ions

There are 2 $PO_4^{3-}$ ions on the right (in $Ca_3(PO_4)_2$), so we put a coefficient of 2 in front of $Na_3PO_4$ on the left.
$2Na_3PO_4 + CaCl_2
ightarrow NaCl + Ca_3(PO_4)_2$

Step 2: Balance Ca atoms

There are 3 Ca atoms on the right (in $Ca_3(PO_4)_2$), so we put a coefficient of 3 in front of $CaCl_2$ on the left.
$2Na_3PO_4 + 3CaCl_2
ightarrow NaCl + Ca_3(PO_4)_2$

Step 3: Balance Na and Cl atoms

On the left, we have 6 Na atoms (from $2Na_3PO_4$) and 6 Cl atoms (from $3CaCl_2$). So we put a coefficient of 6 in front of $NaCl$ on the right.
$2Na_3PO_4 + 3CaCl_2
ightarrow 6NaCl + Ca_3(PO_4)_2$

Problem 16: $\boldsymbol{K + Cl_2

ightarrow KCl}$

Step 1: Balance Cl atoms

There are 2 Cl atoms on the left (in $Cl_2$), so we put a coefficient of 2 in front of $KCl$ on the right.
$K + Cl_2
ightarrow 2KCl$

Step 2: Balance K atoms

Now, we have 2 K atoms on the right, so we put a coefficient of 2 in front of $K$ on the left.
$2K + Cl_2
ightarrow 2KCl$

Problem 17: $\boldsymbol{Al + HCl

ightarrow H_2 + AlCl_3}$

Step 1: Balance Al atoms

There is 1 Al atom on both sides, so Al is balanced for now.

Step 2: Balance Cl atoms

There are 3 Cl atoms on the right (in $AlCl_3$), so we put a coefficient of 3 in front of $HCl$ on the left.
$Al + 3HCl
ightarrow H_2 + AlCl_3$

Step 3: Balance H atoms

On the left, we have 3 H atoms, and on the right, we have 2 H atoms (in $H_2$). The least common multiple of 3 and 2 is 6. So we multiply $HCl$ by 2 and $H_2$ by 3, and adjust Al and $AlCl_3$ accordingly.
$2Al + 6HCl
ightarrow 3H_2 + 2AlCl_3$

Problem 18: $\boldsymbol{N_2 + F_2

ightarrow NF_3}$

Step 1: Balance N atoms

There are 2 N atoms on the left (in $N_2$), so we put a coefficient of 2 in front of $NF_3$ on the right.
$N_2 + F_2
ightarrow 2NF_3$

Step 2: Balance F atoms

On the right, we have 6 F atoms (in $2NF_3$), so we put a coefficient of 3 in front of $F_2$ on the left.
$N_2 + 3F_2
ightarrow 2NF_3$

Problem 19: $\boldsymbol{SO_2 + Li_2Se

ightarrow SSe_2 + Li_2O}$

Step 1: Balance S atoms

There is 1 S atom on both sides, so S is balanced for now.

Step 2: Balance Se atoms

There are 2 Se atoms on the right (in $SSe_2$), so we put a coefficient of 2 in front of $Li_2Se$ on the left.
$SO_2 + 2Li_2Se
ightarrow SSe_2 + Li_2O$

Step 3: Balance Li and O atoms

On the left, we have 4 Li atoms (from $2Li_2Se$) and 2 O atoms (from $SO_2$). On the right, we have 2 Li atoms (in $Li_2O$) and 1 O atom. So we put a coefficient of 2 in front of $Li_2O$ on the right.
$SO_2 + 2Li_2Se
ightarrow SSe_2 + 2Li_2O$

Problem 20: $\boldsymbol{NH_3 + H_2SO_4

ightarrow (NH_4)_2SO_4}$

Step 1: Balance $NH_4^+$ ions

There are 2 $NH_4^+$ ions on the right (in $(NH_4)_2SO_4$), so we put a coefficient of 2 in front of $NH_3$ on the left.
$2NH_3 + H_2SO_4
ightarrow (NH_4)_2SO_4$

Now let's summarize the balanced equations:

1. $\boldsymbol{2AlBr_3 + 3K_2SO_4

ightarrow 6KBr + Al_2(SO_4)_3}$

1. $\boldsymbol{CH_4 + 2O_2

ightarrow CO_2 + 2H_2O}$

1. $\boldsymbol{2Na_3PO_4 + 3CaCl_2

ightarrow 6NaCl + Ca_3(PO_4)_2}$

1. $\boldsymbol{2K + Cl_2

ightarrow 2KCl}$

1. $\boldsymbol{2Al + 6HCl

ightarrow 3H_2 + 2AlCl_3}$

1. $\boldsymbol{N_2 + 3F_2

ightarrow 2NF_3}$

1. $\boldsymbol{SO_2 + 2Li_2Se

ightarrow SSe_2 + 2Li_2O}$

1. $\boldsymbol{2NH_3 + H_2SO_4

ightarrow (NH_4)_2SO_4}$