Question

What is the quantity of gold chloride obtained when 4.5 g gold and 2.1 g chlorine when sealed in a tube and heated at 150 \( ^\circ \)C? (At. masses of Au = 196.97, Cl = 35.45 u)

• A. 4.5 g
• B. 4.8 g
• C. 6.07 g
• D. 20.7 g

Hint:

Always determine the limiting reagent before calculating the product formed.

Answer 1

The Correct Option is C

Step 1: Writing the reaction.
\[ \mathrm{2Au + 3Cl_2 \rightarrow 2AuCl_3} \]

Step 2: Calculating moles of reactants.
\[ \text{Moles of Au} = \frac{4.5}{196.97} \approx 0.0228 \] \[ \text{Moles of Cl}_2 = \frac{2.1}{70.9} \approx 0.0296 \]

Step 3: Identifying limiting reagent.
According to stoichiometry, 2 mol Au require 3 mol Cl\(_2\). For 0.0228 mol Au, required Cl\(_2\) = 0.0342 mol, but only 0.0296 mol is available. Thus, chlorine is the limiting reagent.

Step 4: Calculating moles of AuCl\(_3\).
\[ \text{Moles of AuCl}_3 = \frac{2}{3} \times 0.0296 = 0.0197 \]

Step 5: Calculating mass of AuCl\(_3\).
\[ \text{Mass} = 0.0197 \times (196.97 + 3 \times 35.45) \approx 6.07 \, \text{g} \]

Step 6: Conclusion.
The quantity of gold chloride obtained is 6.07 g.