Question

The reaction of tungsten hexachloride \(\left(\mathrm{WCl}_{6}\right)\) with bismuth gives hexatungsten dodecachloride \(\left(\mathrm{W}_{6} \mathrm{Cl}_{12}\right)\) $$ \mathrm{WCl}_{6}+\mathrm{Bi} \longrightarrow \mathrm{W}_{6} \mathrm{Cl}_{12}+\mathrm{BiCl}_{3} \quad \text { Unbalanced } $$ (a) Balance the equation. (b) How many grams of bismuth react with \(150.0 \mathrm{~g}\) of \(\mathrm{WCl}_{6} ?\) (c) When \(228 \mathrm{~g}\) of \(\mathrm{WCl}_{6}\) react with \(175 \mathrm{~g}\) of \(\mathrm{Bi}\), how much \(\mathrm{W}_{6} \mathrm{Cl}_{12}\) is formed based on the limiting reactant?

Short answer

(a) 6 WCl₆ + 8 Bi → W₆Cl₁₂ + 8 BiCl₃. (b) 105.3 g Bi. (c) 156.0 g W₆Cl₁₂.

Step-by-step solution

Balance the Chemical Equation

To balance the equation, ensure that the number of each type of atom is equal on both sides. The unbalanced equation is: \[ \mathrm{WCl}_{6} + \mathrm{Bi} \rightarrow \mathrm{W}_{6}\mathrm{Cl}_{12} + \mathrm{BiCl}_3 \]Balance the equation step by step:1. First balance tungsten (W): As there are 6 W in \(\mathrm{W}_{6}\mathrm{Cl}_{12}\), place a coefficient 6 in front of \(\mathrm{WCl}_{6}\): \[ 6 \mathrm{WCl}_{6} + \mathrm{Bi} \rightarrow \mathrm{W}_{6}\mathrm{Cl}_{12} + \mathrm{BiCl}_3 \]2. Next, balance chlorine (Cl): 6 \(\mathrm{WCl}_{6}\) produces \(6 \times 6 = 36\) Cl. \(\mathrm{W}_{6}\mathrm{Cl}_{12}\) has 12 Cl, so 36 - 12 = 24 Cl should balance with 8 \(\mathrm{BiCl}_3\) ensuring each \(\mathrm{BiCl}_3\) has 3 Cl. Therefore, add coefficient 8 to \(\mathrm{BiCl}_3\): \[ 6 \mathrm{WCl}_{6} + \mathrm{Bi} \rightarrow \mathrm{W}_{6}\mathrm{Cl}_{12} + 8 \mathrm{BiCl}_3 \]3. Finally, balance bismuth (Bi): As there are 8 Bi in \(8 \mathrm{BiCl}_3\), we need 8 Bi initially: \[ 6 \mathrm{WCl}_{6} + 8 \mathrm{Bi} \rightarrow \mathrm{W}_{6}\mathrm{Cl}_{12} + 8 \mathrm{BiCl}_3 \]Thus, the balanced equation is: \[ 6 \mathrm{WCl}_{6} + 8 \mathrm{Bi} \rightarrow \mathrm{W}_{6}\mathrm{Cl}_{12} + 8 \mathrm{BiCl}_3 \]

Calculate Moles of WCl6

To calculate the moles of \(\mathrm{WCl}_{6}\), determine its molar mass first. The atomic masses are: W = 183.84 g/mol, Cl = 35.45 g/mol.\[\text{Molar Mass of } \mathrm{WCl}_{6} = 183.84 + 6 \times 35.45 = 396.54\text{ g/mol}\]For 150.0 g of \(\mathrm{WCl}_{6}\):\[\text{Moles of } \mathrm{WCl}_{6} = \frac{150.0}{396.54} \approx 0.378\text{ moles} \]

Determine Moles of Bi Needed

The balanced equation shows that 6 moles of \(\mathrm{WCl}_{6}\) require 8 moles of \(\mathrm{Bi}\). For 0.378 moles of \(\mathrm{WCl}_{6}\), calculate \(\mathrm{Bi}\) moles needed using stoichiometry:\[\text{Moles of } \mathrm{Bi} = \frac{8}{6} \times 0.378 \approx 0.504\text{ moles} \]

Calculate Mass of Bi Needed

Determine the mass of \(\mathrm{Bi}\) using its molar mass. The molar mass of Bi is 208.98 g/mol.\[\text{Mass of } \mathrm{Bi} = 0.504 \times 208.98 \approx 105.3 \text{ g} \]

Determine Limiting Reactant for Given Masses

Given 228 g of \(\mathrm{WCl}_{6}\) and 175 g of \(\mathrm{Bi}\), calculate moles for each:\[\text{Moles of } \mathrm{WCl}_{6} = \frac{228}{396.54} \approx 0.575\text{ moles} \]\[\text{Moles of } \mathrm{Bi} = \frac{175}{208.98} \approx 0.837\text{ moles} \]Using stoichiometry (0.575 moles of \(\mathrm{WCl}_{6}\) requires \(\frac{8}{6} \times 0.575 \approx 0.767\text{ moles of } \mathrm{Bi}\)), Bi is excess; therefore, \(\mathrm{WCl}_{6}\) is the limiting reactant.

Calculate W6Cl12 Formed from Limiting Reactant

The balanced equation indicates 6 moles of \(\mathrm{WCl}_{6}\) produce 1 mole of \(\mathrm{W}_{6}\mathrm{Cl}_{12}\). For 0.575 moles of \(\mathrm{WCl}_{6}\):\[\text{Moles of } \mathrm{W}_{6}\mathrm{Cl}_{12} = \frac{1}{6} \times 0.575 \approx 0.0958\text{ moles} \]Calculate the mass using the molar mass of \(\mathrm{W}_{6}\mathrm{Cl}_{12}\) (W = 183.84 x 6, Cl = 35.45 x 12 = 1204.14 + 425.4 = 1629.54 g/mol):\[\text{Mass of } \mathrm{W}_{6}\mathrm{Cl}_{12} = 0.0958 \times 1629.54 \approx 156.0 \text{ g} \]

Key concepts

Limiting Reactant

In chemical reactions, determining the limiting reactant is crucial because it governs the quantity of product formed. A limiting reactant is the substance that is entirely consumed first, stopping the reaction from proceeding because no further reactant can be converted into product.
When you are given the masses of two or more reactants, it's essential to figure out which is the limiting reactant. Here's how you do it:

• Calculate the moles of each reactant using their respective molar masses.
• Use the balanced chemical equation to find the stoichiometric ratio between the reactants.
• Compare these mole ratios to identify which reactant is present in less than the required stoichiometric proportion.

In our exercise above, we saw that when reacting 228 g of \(\mathrm{WCl}_{6}\) with 175 g of \(\mathrm{Bi}\), the limiting reactant is \(\mathrm{WCl}_{6}\). Although \(\mathrm{Bi}\) is present in a larger amount, the smaller amount of \(\mathrm{WCl}_{6}\) restricts the amount of product \(\mathrm{W}_{6}\mathrm{Cl}_{12}\) that can be formed. Calculating this ensures you use resources efficiently and predict yields accurately.

Stoichiometry

Stoichiometry is the study of the quantitative relationships, or ratios, between the reactants and products in chemical reactions. It involves using a balanced chemical equation to find the amount of reactants needed or products formed.
The foundation of stoichiometry is based on the Law of Conservation of Mass, which states that in a chemical reaction, matter is neither created nor destroyed. This concept implies that you must have the same number of each type of atom on both sides of a chemical equation.To perform stoichiometric calculations:

• First, balance your chemical equation to ensure it reflects the conservation of mass.
• Use the coefficients from the balanced equation to determine the mole ratios that convert between reactants and products.
• Apply these ratios to calculate the number of moles of each required substance.

In the example provided, we used stoichiometry to find out that 6 moles of \(\mathrm{WCl}_{6}\) correspond to 8 moles of \(\mathrm{Bi}\). When you know this ratio, you can calculate how much of each reactant is needed and how much product can form, essential for both lab planning and industrial chemical production.

Mole Calculations

Mole calculations are an integral component of understanding chemical reactions as they quantify the amount of a substance. The mole is a unit that represents 6.022 x 10\(^{23}\) entities (atoms, molecules, etc.) of a material.
To perform mole calculations, you need the following steps:

• Identify the number of moles from a given weight using the formula: \( \text{Moles} = \frac{\text{mass}}{\text{molar mass}} \).
• The molar mass is calculated by adding up the atomic masses of all atoms in a molecule.
• Use the moles obtained to engage in stoichiometric calculations or to determine reagents/products' masses if needed.

For example, in our exercise, to find the number of moles of \(\mathrm{WCl}_6\) from a given mass, we calculated its molar mass as 396.54 g/mol. From 150 g of \(\mathrm{WCl}_6\), we find \(0.378\) moles using the formula provided. Such calculations are pivotal in determining reaction scales and converting between mass and mole assessments, thus guiding primary chemical analysis and predictions.