Potassium oxide is an ionic compound. The potassium has a charge of <span>K+</span> and oxygen has a charge of <span>O<span>2−</span></span>. We need 2 potassium ions to balance one oxide ion making the formula <span><span>K2</span>O</span>.
Potassium hydroxide is an ionic compound. The potassium has a charge of <span>K+</span> and hydroxide has a charge of <span>OH−</span>. We need 1 potassium ion to balance one hydroxide ion making the formula KOH.
<span><span>K2</span>O+<span> H2</span>O→KOH</span>
To balance the equation we place a coefficient of 2 in front of the potassium hydroxide.
<span><span>K2</span>O+<span>H2</span>O→2KOH</span>
I hope this was helpful.
Barium :
with +2 being the charge
Oxygen :
with -2 being the charge
The given equation can be written as:
Ba + O = BaO
Since the sum charges of Barium and Oxygen equals 0, there is no need to add subscripts.
Both Ba and O appear on the left and right side of the equation once, so there is no need to add a coefficient.
Ba + O = BaO is balanced
Answer:
[Zn²⁺] = 4.78x10⁻¹⁰M
Explanation:
Based on the reaction:
ZnBr₂(aq) + K₂CO₃(aq) → ZnCO₃(s) + 2KBr(aq)
The zinc added produce the insoluble ZnCO₃ with Ksp = 1.46x10⁻¹⁰:
1.46x10⁻¹⁰ = [Zn²⁺] [CO₃²⁻]
We can find the moles of ZnBr₂ added = Moles of Zn²⁺ and moles of K₂CO₃ = Moles of CO₃²⁻ to find the moles of CO₃²⁻ that remains in solution, thus:
<em>Moles ZnB₂ (Molar mass: 225.2g/mol) = Moles Zn²⁺:</em>
6.63g ZnBr₂ * (1mol / 225.2g) = 0.02944moles Zn²⁺
<em>Moles K₂CO₃ = Moles CO₃²⁻:</em>
0.100L * (0.60mol/L) = 0.060 moles CO₃²⁻
Moles CO₃²⁻ in excess: 0.0600moles CO₃²⁻ - 0.02944moles =
0.03056moles CO₃²⁻ / 0.100L = 0.3056M = [CO₃²⁻]
Replacing in Ksp expression:
1.46x10⁻¹⁰ = [Zn²⁺] [0.3056M]
<h3>[Zn²⁺] = 4.78x10⁻¹⁰M</h3>
