25.0 ml of 0.235 M NiCl2 are combined with 30.0 ml of 2.260 M KOH. How many grams of nickel hydroxide will precipitate?
To find out how many grams of nickel hydroxide will precipitate, we need to use the concept of stoichiometry and the balanced chemical equation.
The balanced chemical equation for the reaction between nickel chloride (NiCl2) and potassium hydroxide (KOH) is:
NiCl2 + 2KOH -> Ni(OH)2 + 2KCl
In this equation, we can see that 1 mole of nickel chloride reacts with 2 moles of potassium hydroxide to produce 1 mole of nickel hydroxide and 2 moles of potassium chloride.
First, we need to calculate the number of moles of NiCl2 and KOH in the given volumes.
Moles of NiCl2 = volume (in L) * concentration (in mol/L)
= 0.025 L * 0.235 mol/L
= 0.005875 mol
Moles of KOH = volume (in L) * concentration (in mol/L)
= 0.03 L * 2.260 mol/L
= 0.0678 mol
Now, we can compare the moles of reactants based on the stoichiometry of the equation.
From the balanced equation, we see that 1 mole of NiCl2 reacts with 2 moles of KOH to produce 1 mole of Ni(OH)2. Therefore, the limiting reactant is NiCl2 since we have less moles of NiCl2 than 2 moles of KOH.
Based on the stoichiometry, 1 mole of NiCl2 produces 1 mole of Ni(OH)2. So, the moles of Ni(OH)2 formed would also be 0.005875 mol.
To convert moles of Ni(OH)2 to grams, we need to use its molar mass.
Molar mass of Ni(OH)2 = atomic mass of Ni + 2 * (atomic mass of O + atomic mass of H)
= 58.69 g/mol + 2 * (16.00 g/mol + 1.01 g/mol)
= 92.69 g/mol
Now, we can calculate the grams of nickel hydroxide formed:
Grams of Ni(OH)2 = moles of Ni(OH)2 * molar mass of Ni(OH)2
= 0.005875 mol * 92.69 g/mol
= 0.545 g
Therefore, approximately 0.545 grams of nickel hydroxide will precipitate.
NiCl2 + 2KOH ==> Ni(OH)2 + 2KCl
mols NiCl2 = M x L = approx 0.006 but you need a more accurate answer.
mols KOH = M x L = approx 0.07--again you need a more accurate answer.
mols Ni(OH)2 formed if all NiCl2 used will be 0.006 x 1 mol (Ni(OH)2/1 mol NiCl2) = approx 0.006
mols Ni(OH)2 formed if all KOH used will be 0.07 x (Ni(OH)2/2KOH) = approx 0.035
You see the answers are not the same; the correct value in limiting reagent (LR) problems is ALWAYS the smaller value and the reagent producing that value is the LR.
g Ni(OH)2 formed = mols Ni(OH)2 x molar mass Ni(OH)2