Answer:
a. CaCO₃ (s) → CaO (s) + CO₂(g) ΔH
b. 2C₄H₁₀ (g) + 13O₂(g) → 8CO₂ (g) + 10H₂O(g)
c. MgCl₂ (aq) + 2NaOH(aq) → Mg(OH)₂ (s) ↓ + 2NaCl (aq)
d. 2H₂O(g) + 2Na(s) → 2NaOH(s) + H₂(g)
Explanation:
We identify the compound by the formula and we make the reaction with the phases of each reactant and product:
a. CaCO₃ (s); CaO (s), CO₂(g)
The reaction is: CaCO₃ (s) → CaO (s) + CO₂(g) ΔH
We add the ΔH symbol because there is heat added to the reaction, which means a enthalpy change
b. C₄H₁₀ (g), O₂(g) ; CO₂ (g), H₂O(g)
The reaction is: 2C₄H₁₀ (g) + 13O₂(g) → 8CO₂ (g) + 10H₂O(g)
In order to balance, we add 10 moles to water, so in total we have 20 H. Therefore we need 2 moles of butane to have 20 H, but we also have 8 C. So we need 8 C for the CO₂. By the end we have 16 O in CO₂ plus 10 O from water. In conclussion we need 13 O₂ in the reactant side to get the final balance
c. MgCl₂ (aq), NaOH(aq) ; Mg(OH)₂ (aq) , NaCl
The reaction is: MgCl₂ (aq) + 2NaOH(aq) → Mg(OH)₂ (s) ↓ + 2NaCl (aq)
As the Mg(OH)₂ is solid, it indicates that we made a precipitate in the reaction (That's why we add ↓, in the product side). We need 2NaCl, to balance the Cl and 2 NaOH to balance the Na.
Finally the Mg is balanced
d. H₂O(g), Na(s) ; NaOH (s), H₂(g)
Reaction is: 2H₂O(g) + 2Na(s) → 2NaOH(s) + H₂(g)
We add 2 water in order to balance the hydrogen gas, but we need a 2 in the hydoxide. Finally we will balance the Na with also a 2.
