The balanced equation for the above reaction is as follows;
C + H₂O ---> H₂ + CO
stoichiometry of C to H₂O is 1:1
1 mol of C reacts with 1 mol of H₂O
we need to find which is the limiting reactant
2 mol of C and 3.1 mol of H₂O
therefore C is the limiting reactant and H₂O is in excess.
stoichiometry of C to H₂ is 1:1
then number of H₂ moles formed are equal to C moles reacted
number of H₂ moles formed = 2 mol
Molar solubility is the number of moles that are dissolved in 1 L solution.
when BaF₂ dissolves it dissociates into the following ions
BaF₂ --> Ba²⁺ + 2F⁻
if the molar solubility of BaF₂ is X, then molar solubility of Ba²⁺ is X and F⁻ is 2x
then the formula for the solubility product constant -ksp is;
ksp = [Ba²⁺][F⁻]²
ksp = X * (2X)²
ksp = 4X³
since ksp = 1.7 x 10⁻⁶
4X³ = 1.7 x 10⁻⁶
X = 0.0075 M
molar solubility of BaF₂ is 0.0075 M