Not sure how to solve.

Question

If 28.0 g of NaOH is added to 0.550 L of 1.00 M Ni(NO₃)₂, how many grams of Ni(OH)₂ will be formed in the following precipitation reaction? 2 NaOH(aq) + Ni(NO₃)₂(aq) → Ni(OH)₂ (s) + 2 NaNO₃ (aq).

Answer 1

Ni(NO₃)₂ + 2NaOH ==> 2NaNO₃ + Ni(OH)₂(s) ... balanced equation

Since we are given the amounts of both reactants, we must find the limiting reactant. An easy way to do that is to divide the moles of each reactant by the corresponding coefficient in the balanced equation. Whichever value is less is the limiting reactant.

For NaOH:= 28.0 g x 1 mol / 40.0 g = 0.70 moles NaOH (÷2->0.35)

For Ni(NO₃)₂ = 0.550 L x 1.0 mol/L = 0.550 moles Ni(NO₃)₂ (÷1->0.55)

So, NaOH is limiting and will determine the mass of Ni(OH)₂ that can form.

0.70 mols NaOH x 1 mol Ni(OH)₂ / 2 mols NaOH x 92.7 g Ni(OH)₂ / mole = 32.4 g Ni(OH)₂

Answer 2

General:

1. Convert mass (g) NaOH ⇒ mol NaOH.
2. Convert Molarity (M) of Ni(NO₃)₂ ⇒ mol Ni(NO₃)₂.
3. Convert both answers in steps 1 and 2 ⇒ mol Ni(OH)₂.
4. Take the LOWER mol number (limiting reagent) and convert to mass (g) Ni(OH)₂.

The goal is to get the product into moles so you can convert to mass using moles of the limiting reactant.

The reagents may be given in different ways in similar word problems:

1. Moles - simplest to solve, just a molar conversion step.
2. Mass - convert mass to moles using molecular mass of each reagent.
3. Molarity and Volume - convert molarity and volume to moles of reagent solution using m = M x V.
4. Pressure, Volume, and Temperature - convert P,V, & T using the ideal gas law, (PxV) ÷ (RxT).
5. Basically anything that can give you moles can be used.

Example: 14g NaOH and 0.225L of 1M Ni(NO3)2 to g Ni(OH)2 [similar steps to solve the problem]=>

1) Convert mass of NaOH to moles of NaOH.

mass NaOH [g] / MW NaOH [mol/g] = mol NaOH

14g NaOH / 39.997g/mol [molecular weight] =

= 0.350 mol NaOH

2) volume Ni(NO3)2 x Molarity of Ni(NO3)2 = mol Ni(NO3)2

0.225 L Ni(NO3)2 x 1 M Ni(NO3)2 = 0.225 L Ni(NO3)2 x 1 mol/L Ni(NO3)2 =

= 0.255 mol Ni(NO3)2

3) Convert both to moles of Ni(OH)2.

This will tell you the MAXIMUM and MINNIMUM that can be produced.

0.350 mol NaOH x (1 mol Ni(OH)2 / 2 mol NaOH) =

0.175 mol Ni(OH)2

0.255 mol Ni(NO3)2 x (1 mol Ni(NO3)2 / 1 mol Ni(OH)2) =

0.255 mol Ni(OH)2

4) Since it is impossible to make more than you put in (limiting reagent), pick the lowest number of moles that can be produced with the given starting products.

In this case, NaOH (0.175 mol Ni(OH)2 produced).

5) Convert moles Ni(OH)2 to grams using molecular weight of Ni(OH)2.

0.175 mol Ni(OH)2 x 92.707 g Ni(OH)2/mol Ni(OH)2 =

= 16.2 g Ni(OH)2

The same math works for your problem. I know it's a long explanation, but it's every step.