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