Ni₂(SO₄)₃ + HCl Reaction: Moles & NiCl₃ Yield Calculation

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Hey guys! Let's break down this chemistry problem step-by-step. We're diving into a reaction involving Nickel(III) sulfate and hydrochloric acid, and we'll be figuring out molecular weights, moles, and product yield. Grab your calculators, and let's get started!

a) Calculating Molecular Weights

First up, we need to calculate the molecular weights of all the compounds involved in the reaction. This is a crucial first step because it allows us to convert between grams and moles, which is essential for stoichiometry. Remember, the molecular weight of a compound is the sum of the atomic weights of all the atoms in the molecule. You can find the atomic weights on the periodic table.

Let's break it down for each compound:

  • Ni₂(SO₄)₃ (Nickel(III) sulfate):

    • We have 2 Nickel (Ni) atoms, each with an atomic weight of approximately 58.69 g/mol. So, 2 * 58.69 = 117.38 g/mol.
    • Next, we have 3 Sulfate (SO₄) groups. Each sulfate group contains 1 Sulfur (S) atom (atomic weight ~32.06 g/mol) and 4 Oxygen (O) atoms (atomic weight ~16.00 g/mol). So, for one SO₄ group: 32.06 + (4 * 16.00) = 96.06 g/mol.
    • Since we have 3 sulfate groups, the total contribution from sulfate is 3 * 96.06 = 288.18 g/mol.
    • Adding the contributions from Nickel and sulfate, the molecular weight of Ni₂(SO₄)₃ is 117.38 + 288.18 = 405.56 g/mol.

  • HCl (Hydrochloric acid):

    • This one's simpler. We have 1 Hydrogen (H) atom (atomic weight ~1.01 g/mol) and 1 Chlorine (Cl) atom (atomic weight ~35.45 g/mol).
    • So, the molecular weight of HCl is 1.01 + 35.45 = 36.46 g/mol.

  • NiCl₃ (Nickel(III) chloride):

    • We have 1 Nickel (Ni) atom (atomic weight ~58.69 g/mol) and 3 Chlorine (Cl) atoms (atomic weight ~35.45 g/mol).
    • The molecular weight of NiCl₃ is 58.69 + (3 * 35.45) = 58.69 + 106.35 = 165.04 g/mol.

  • H₂SO₄ (Sulfuric acid):

    • We have 2 Hydrogen (H) atoms (atomic weight ~1.01 g/mol), 1 Sulfur (S) atom (atomic weight ~32.06 g/mol), and 4 Oxygen (O) atoms (atomic weight ~16.00 g/mol).
    • The molecular weight of H₂SO₄ is (2 * 1.01) + 32.06 + (4 * 16.00) = 2.02 + 32.06 + 64.00 = 98.08 g/mol.

So, to recap, the molecular weights are:

  • Ni₂(SO₄)₃: 405.56 g/mol
  • HCl: 36.46 g/mol
  • NiCl₃: 165.04 g/mol
  • H₂SO₄: 98.08 g/mol

b) Calculating Moles from 0.156 g of Each Reactant

Now that we have the molecular weights, we can calculate how many moles are present in 0.156 g of each starting material. Remember, the number of moles is calculated by dividing the mass (in grams) by the molecular weight (in g/mol).

The formula is:

Moles = Mass (g) / Molecular Weight (g/mol)

Let's do the calculations:

  • Ni₂(SO₄)₃:

    • Moles of Ni₂(SO₄)₃ = 0.156 g / 405.56 g/mol = 0.000385 moles (approximately)

  • HCl:

    • Moles of HCl = 0.156 g / 36.46 g/mol = 0.00428 moles (approximately)

So, if we use 0.156 g of each starting material, we're dealing with approximately 0.000385 moles of Ni₂(SO₄)₃ and 0.00428 moles of HCl.

c) Calculating Grams of NiCl₃ Produced

This is where things get a bit more complex, but don't worry, we'll take it slow. To figure out how many grams of NiCl₃ can be made, we first need to consider the balanced chemical equation for the reaction.

The unbalanced equation is:

Ni₂(SO₄)₃(aq) + HCl(aq) → NiCl₃(s) + H₂SO₄(aq)

Let's balance it! Balancing chemical equations ensures that we have the same number of each type of atom on both sides of the equation, which is essential for accurate stoichiometric calculations.

The balanced equation is:

Ni₂(SO₄)₃(aq) + 6 HCl(aq) → 2 NiCl₃(s) + 3 H₂SO₄(aq)

Now we can see the molar ratios between the reactants and products. The balanced equation tells us that 1 mole of Ni₂(SO₄)₃ reacts with 6 moles of HCl to produce 2 moles of NiCl₃ and 3 moles of H₂SO₄.

1. Identify the Limiting Reactant

The first thing we need to do is figure out which reactant is the limiting reactant. The limiting reactant is the one that is completely consumed in the reaction, and it determines the maximum amount of product that can be formed. To find the limiting reactant, we compare the mole ratio of the reactants to the stoichiometric ratio from the balanced equation.

We have 0.000385 moles of Ni₂(SO₄)₃ and 0.00428 moles of HCl. According to the balanced equation, 1 mole of Ni₂(SO₄)₃ reacts with 6 moles of HCl. So, we need to see if we have enough HCl to react with all the Ni₂(SO₄)₃.

  • Moles of HCl needed to react with 0.000385 moles of Ni₂(SO₄)₃ = 0.000385 moles * 6 = 0.00231 moles

We have 0.00428 moles of HCl, which is more than the 0.00231 moles needed. This means that Ni₂(SO₄)₃ is the limiting reactant because we'll run out of it before we run out of HCl.

2. Calculate Moles of NiCl₃ Produced

Now that we know Ni₂(SO₄)₃ is the limiting reactant, we can calculate how many moles of NiCl₃ will be produced. From the balanced equation, 1 mole of Ni₂(SO₄)₃ produces 2 moles of NiCl₃.

  • Moles of NiCl₃ produced = 0.000385 moles Ni₂(SO₄)₃ * (2 moles NiCl₃ / 1 mole Ni₂(SO₄)₃) = 0.00077 moles NiCl₃

3. Calculate Grams of NiCl₃ Produced

Finally, we can convert moles of NiCl₃ to grams using the molecular weight of NiCl₃ (165.04 g/mol).

  • Grams of NiCl₃ produced = 0.00077 moles * 165.04 g/mol = 0.127 grams (approximately)

So, from 0.156 g of each starting material, we can produce approximately 0.127 grams of NiCl₃.

In Conclusion

We've tackled a full stoichiometry problem, guys! We calculated molecular weights, converted grams to moles, balanced a chemical equation, identified the limiting reactant, and finally, calculated the theoretical yield of NiCl₃. These are fundamental skills in chemistry, and mastering them will help you solve all sorts of problems. Keep practicing, and you'll become a stoichiometry pro in no time!