CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,
CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Initial Y - -
Change -X +X +X
Equilibrium Y-X X X
Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²
Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
X = 5.79 x 10⁻⁵ M
Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
= 5.79 x 10⁻⁵ mol/L
Molar mass of CaCO₃ = 100 g mol⁻¹
Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
= 5.79 x 10⁻³ g/L
No is the answer your welcome well I don’t.
Answer:
10.85 g of water
Explanation:
First we write the balanced chemical equation

Then we calculate the number of moles of nitric acid produced
n(HNO3) = 
According to the balanced equation, water needed in moles is always half the number of moles of HNO3 produced. So since we will produce 1.2044 mol of HNO3, we will need 0.6022 mol of water. Now to calculate what mass that is:
mass(water)=number of moles*molar mass=0.6022mol*18.02g/mol=10.85g
When the balanced reaction equation is:
2HCl(aq) + Ca(OH)2(aq) → CaCl2(aq) + 2H2O(l)
from the balanced equation, we can get the molar ratio between HCl & Ca(OH)2
2:1
∴ the volume of Ca(OH)2 = 15.8 L HCl * 1.51 m HCl * (1mol Ca(OH)2/ 2mol HCl) * (1L ca(OH)2/0.585 mol Ca(OH)2
= 20.4 L