Question

na + cl₂ => nacl
rb + s₈ => rb₂s
h₃po₄ + ca(oh)₂ => ca₃(po₄)₂ + h₂o
nh₃ + hcl => nh₄cl
li + h₂o => lioh + h₂
ca₃(po₄)₂ + sio₂ + c => casio₃ + co + p
nh₃ + o₂ => n₂ + h₂o
fes₂ + o₂ => fe₂o₃ + so₂

Explanation:

Step1: Balance the first equation

The un - balanced equation is $Na + Cl_2
ightarrow NaCl$. There are 2 chlorine atoms on the left - hand side and 1 on the right - hand side. To balance the chlorine atoms, we put a 2 in front of $NaCl$. Then to balance the sodium atoms, we put a 2 in front of $Na$. The balanced equation is $2Na + Cl_2 = 2NaCl$.

Step2: Balance the second equation

The un - balanced equation is $Rb+S_8
ightarrow Rb_2S$. There are 8 sulfur atoms on the left - hand side. To balance the sulfur atoms, we put an 8 in front of $Rb_2S$. Then to balance the rubidium atoms, we put a 16 in front of $Rb$. The balanced equation is $16Rb + S_8=8Rb_2S$.

Step3: Balance the third equation

The un - balanced equation is $H_3PO_4+Ca(OH)_2
ightarrow Ca_3(PO_4)_2 + H_2O$. There are 2 phosphate groups in $Ca_3(PO_4)_2$, so we put a 2 in front of $H_3PO_4$. Then, since there are 3 calcium atoms in $Ca_3(PO_4)_2$, we put a 3 in front of $Ca(OH)_2$. Counting the hydrogen and oxygen atoms, we find that there are 12 hydrogen atoms and 14 oxygen atoms on the left - hand side. On the right - hand side, we need to put a 6 in front of $H_2O$. The balanced equation is $2H_3PO_4 + 3Ca(OH)_2=Ca_3(PO_4)_2+6H_2O$.

Step4: Balance the fourth equation

The un - balanced equation is $NH_3+HCl
ightarrow NH_4Cl$. This equation is already balanced as there is 1 nitrogen, 4 hydrogen, and 1 chlorine atom on both sides. The balanced equation is $NH_3 + HCl=NH_4Cl$.

Step5: Balance the fifth equation

The un - balanced equation is $Li + H_2O
ightarrow LiOH+H_2$. There are 2 hydrogen atoms in $H_2$ on the right - hand side. To balance the hydrogen atoms, we put a 2 in front of $Li$ and a 2 in front of $LiOH$ and a 2 in front of $H_2O$. The balanced equation is $2Li + 2H_2O = 2LiOH+H_2$.

Step6: Balance the sixth equation

The un - balanced equation is $Ca_3(PO_4)_2+SiO_2 + C
ightarrow CaSiO_3+CO + P$. First, to balance the calcium atoms, we put a 3 in front of $CaSiO_3$. Then, to balance the phosphate groups, we note that there are 2 phosphate groups in $Ca_3(PO_4)_2$, so we need to get 2 phosphorus atoms on the right - hand side. To balance the oxygen atoms, we consider the oxygen in $Ca_3(PO_4)_2$, $SiO_2$ and the products. After some trial and error, the balanced equation is $Ca_3(PO_4)_2+3SiO_2 + 5C=3CaSiO_3+5CO + 2P$.

Step7: Balance the seventh equation

The un - balanced equation is $NH_3+O_2
ightarrow N_2+H_2O$. First, to balance the nitrogen atoms, we put a 2 in front of $NH_3$. Then, to balance the hydrogen atoms, since there are 6 hydrogen atoms in $2NH_3$, we put a 3 in front of $H_2O$. To balance the oxygen atoms, we note that there are 3 oxygen atoms in $3H_2O$, so we need $\frac{3}{2}$ in front of $O_2$. To get whole - number coefficients, we multiply all coefficients by 2. The balanced equation is $4NH_3+3O_2 = 2N_2+6H_2O$.

Step8: Balance the eighth equation

The un - balanced equation is $FeS_2+O_2
ightarrow Fe_2O_3+SO_2$. First, to balance the iron atoms, we put a 2 in front of $FeS_2$. Then, since there are 4 sulfur atoms in $2FeS_2$, we put a 4 in front of $SO_2$. Counting the oxygen atoms, there are 11 oxygen atoms on the right - hand side, so we put $\frac{11}{2}$ in front of $O_2$. To get whole - number coefficients, we multiply all coefficients by 2. The balanced equation is $4FeS_2+11O_2 = 2Fe_2O_3+8SO_2$.

Answer:

$2Na + Cl_2 = 2NaCl$; $16Rb + S_8=8Rb_2S$; $2H_3PO_4 + 3Ca(OH)_2=Ca_3(PO_4)_2+6H_2O$; $NH_3 + HCl=NH_4Cl$; $2Li + 2H_2O = 2LiOH+H_2$; $Ca_3(PO_4)_2+3SiO_2 + 5C=3CaSiO_3+5CO + 2P$; $4NH_3+3O_2 = 2N_2+6H_2O$; $4FeS_2+11O_2 = 2Fe_2O_3+8SO_2$