A solution of 118 mL of 0.160 M KOH is mixed with a solution of 240 mL of 0.230 M NiSO4.
How many grams of this precipitate (Ni(OH)2) form?
I have the balanced equation as 2KOH (aq)+ NiSO4 (aq) ----> K2SO4 (aq) + Ni(OH)2
I don't need the answer but I would really like to know how to do it.
To determine the amount of precipitate (Ni(OH)2) formed in this reaction, you need to use stoichiometry to find the limiting reactant and then calculate the moles of precipitate formed based on the balanced equation.
1. Start by calculating the moles of each reactant:
Moles of KOH = Volume (in liters) x concentration
= 0.118 L x 0.160 mol/L
= 0.01888 mol
Moles of NiSO4 = Volume (in liters) x concentration
= 0.240 L x 0.230 mol/L
= 0.0552 mol
2. Now, use the stoichiometry of the balanced equation to determine the moles of precipitate formed. According to the equation, 2 moles of KOH react with 1 mole of NiSO4 to form 1 mole of Ni(OH)2.
Since the stoichiometric ratio is 2:1, the limiting reactant will be completely consumed, and the other reactant will be in excess.
In this case, the NiSO4 is the limiting reactant because its moles (0.0552 mol) are smaller than those of KOH (0.01888 mol).
3. Calculate the moles of Ni(OH)2 formed based on the limiting reactant:
Moles of Ni(OH)2 = Moles of limiting reactant (NiSO4) x stoichiometric ratio
= 0.0552 mol x 1 mol Ni(OH)2 / 1 mol NiSO4
= 0.0552 mol
4. Finally, calculate the mass of Ni(OH)2 formed using its molar mass:
Mass of Ni(OH)2 = Moles of Ni(OH)2 x molar mass of Ni(OH)2
= 0.0552 mol x (58.7 g/mol + 2 * 1.01 g/mol + 2 * 16.00 g/mol)
= 4.98 g
Therefore, approximately 4.98 grams of Ni(OH)2 will form in this reaction.
118mL of 0.16M gives 0.118 * 0.16 = 0.01888 moles of KOH
Now the equation tells you how many moles of Ni(OH)2 you will get
convert that to grams.