H2so4 + pb(oh)4 = pb(so4)2 + h2o
fe2o3(s)+ co(g)= fe(i)+ co2(g)
ca3(po4)2 + h2so4 = caso4 + ca(h2po4)2
co(g)+ h2(g)= c8h18(i)+ h2o

solve one of them fasttt with work
balancing equations

2H₂SO₄ + Pb(OH)₄  →  Pb(SO₄)₂   +  4H₂O

Fe₂O₃(s)  + 3CO(g)  →  2Fe (I)  + 3CO₂(g)

Ca₃(PO₄)₂  +  2H₂SO₄  →  2CaSO₄  +  Ca(H₂PO₄)₂

8CO(g)  +   17H₂ (g)  →  C₈H₁₈ (I)  +   8H₂O

Explanation:

H₂SO₄ + Pb(OH)₄  →  Pb(SO₄)₂   +   H₂O

- Look that in product side we have 2 sulfate, so we need 2 sulfate in the sulfuric acid, in reactant side.

2H₂SO₄ + Pb(OH)₄  →  Pb(SO₄)₂   +   H₂O

Now we have 2 protons in the acid (H⁺), and we have 2 mol of acid, so in totally we have 4 H, and there are 4 H in the  lead hydroxide (8H); in reactant side we have 8 H, so we must complete with 4, the water.

2H₂SO₄ + Pb(OH)₄  →  Pb(SO₄)₂   +   4H₂O

1 Pb, 2S, 12 O and 8 H in both side - BALANCED

Fe₂O₃(s)  + CO(g)  → Fe (I)  + CO₂(g)

We have 2 Fe in reactant side, so we must add a 2 to Fe, in product side.

Fe₂O₃(s)  + CO(g)  →  2Fe (I)  + CO₂(g)

In reactant side, we have 4 oxygens and we have 2, in product side so if we add a 2 at CO₂, we are also modifying the amount of C, so we must also add a 2 in CO from reactant side

Fe₂O₃(s)  + 2CO(g)  →  2Fe (I)  + 2CO₂(g)

We disbalance the O now, so we must add 3 which is the perfect choice .

We have 3 carbons each side, and we have 3 O from Fe₂O₃ + 3 O in CO, and in product side we also have in totally 6 atoms of O

Fe₂O₃(s)  + 3CO(g)  →  2Fe (I)  + 3CO₂(g)

BALANCED

Ca₃(PO₄)₂ + H₂SO₄  →  CaSO₄  +  Ca(H₂PO₄)₂

We have 3 Ca in reactant side, so we add 2 Ca to product side, as we have 1 Ca in the Ca(H₂PO₄)₂ . At the same time, we add 2 SO₄ in the sulfuric acid from reactant side

Ca₃(PO₄)₂  +  2H₂SO₄  →  2CaSO₄  +  Ca(H₂PO₄)₂  -  BALANCED

CO(g)  +   H₂ (g)  →  C₈H₁₈ (I)  +   H₂O

We have 8 C in product side, so we add 8C in CO in reactant side.

8CO(g)  +   H₂ (g)  →  C₈H₁₈ (I)  +   H₂O

As we added 8, we modified amount of oxygen, so now we must add 8 in H₂O. In reactant side we have, (16 + 18) 34 hydrogens, so we must add a 17 in reactant side.  

8CO(g)  +   17H₂ (g)  →  C₈H₁₈ (I)  +   8H₂O - BALANCED

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.