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

In order to solve this problem follow these steps:

(1) Is your equation is balanced? (yes, balanced)

(2) convert grams of NaOH to moles of NaOH using molar mass

(3) use mole ratios to go from NaOH to g of Ni(OH)₂

(4) use molar mass to convert moles of Ni(OH)₂ to grams of Ni(OH)₂

Worked:

1. 28.0 g NaOH (1 mol NaOH/40.01 g NaOH) = 0.6998 moles of NaOH
2. according to the balanced equation there are 1 moles of NaOH per 1 mole of Ni(OH)₂
3. 0.6698 mol NaOH * (1 mol Ni(OH)₂ / 2 mol NaOH)
4. 0.335 moles of Ni(OH)₂
5. 0.0335 mol Ni(OH)₂ *(92.71 g/1 mol) =
6. final answer: 3.11 grams Ni(OH)₂