Answer:
The balanced equation for the reaction is:
Pb(ClO₃)₂ (aq) + 2NaI (aq) → PbI₂(s)↓ + 2NaClO₃
Explanation:
The reaction of Pb²⁺ with I⁻ makes a precipitate of lead iodide
Answer:
Pb(ClO3)2(aq) + 2NaI(aq) → PbI2(s) + 2Na(ClO3)(aq)
Explanation:
Step 1: Data given
Reactants are: Pb(ClO3)2 and NaI
Step 2: Balancing the equation
Pb(ClO3)2(aq) + NaI(aq) → PbI(s) + NaClO3(aq)
On the left side, we have 2x Cl, on the right side we have 1x Cl
To balance the amount of Cl, we have to multiply Na(ClO3)3 by 2
Pb(ClO3)2(aq) + NaI(aq) → PbI2(s) + 2Na(ClO3)(aq)
On the right side we have 2x Na, on the left side we have 1x Na.
To balance the amount of Na, we have to multiply NaI (on the left side) by 2
Now the equation is balanced
Pb(ClO3)2(aq) + 2NaI(aq) → PbI2(s) + 2Na(ClO3)(aq)
