Answer:
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
