A student takes 26.0 grams of Nickel II chloride and mixes it with 10.0 grams of sodium hydroxide. How much nickel II hydroxide will be produced given the student has a 4.80% error?
4.80% error in what?
Technically, no niclel(II) hydroxide will be formed by mixing two solids.
To determine how much nickel II hydroxide will be produced, we need to calculate the theoretical yield and then consider the 4.80% error.
Here's how you can calculate it step-by-step:
Step 1: Determine the balanced chemical equation:
The balanced equation for the reaction between nickel II chloride (NiCl2) and sodium hydroxide (NaOH) is:
NiCl2 + 2NaOH → Ni(OH)2 + 2NaCl
Step 2: Calculate the molar masses:
- Molar mass of NiCl2 = 58.7 g/mol (28.7 g/mol for Ni + 2*35.5 g/mol for Cl)
- Molar mass of NaOH = 40.0 g/mol (22.9 g/mol for Na + 16.0 g/mol for O)
- Molar mass of Ni(OH)2 = 74.7 g/mol (2*28.7 g/mol for Ni + 2*16.0 g/mol for O + 2*1.0 g/mol for H)
Step 3: Convert the given masses to moles:
- Moles of NiCl2 = mass (g) / molar mass (g/mol) = 26.0 g / 58.7 g/mol ≈ 0.442 mol
- Moles of NaOH = mass (g) / molar mass (g/mol) = 10.0 g / 40.0 g/mol = 0.250 mol
Step 4: Identify the limiting reagent:
The reactant that is completely consumed in the reaction is the limiting reagent. To determine the limiting reagent, compare the mole ratio between NiCl2 and NaOH in the balanced equation (1:2). Since the mole ratio is 1:2, we have twice as many moles of NaOH as NiCl2. Therefore, NiCl2 is the limiting reagent since it's available in lesser moles.
Step 5: Calculate the theoretical yield:
The theoretical yield is the maximum amount of product that can be obtained based on the limiting reagent. In this case, the limiting reagent is NiCl2, so the moles of Ni(OH)2 produced will be equal to the moles of NiCl2 used.
- Moles of Ni(OH)2 = Moles of NiCl2 = 0.442 mol
Step 6: Convert moles of Ni(OH)2 to grams:
- Mass (g) of Ni(OH)2 = moles of Ni(OH)2 * molar mass (g/mol) = 0.442 mol * 74.7 g/mol = 33.0 g
So, the theoretical yield of Ni(OH)2 is 33.0 grams.
Step 7: Consider the 4.80% error:
To determine the amount of Ni(OH)2 produced with a 4.80% error, we need to calculate 4.80% of the theoretical yield.
- 4.80% of 33.0 g = 0.048 * 33.0 g ≈ 1.59 g
Therefore, the nickel II hydroxide produced with a 4.80% error is approximately 1.59 grams.
Note: The error can be applied in different ways, such as subtracting it from or adding it to the theoretical yield, depending on the context or experimental setup.