Question

zn + 2hbr → znbr₂ + h₂
32.7 g of solid zinc was dropped into excess hydrobromic acid.
what is the theoretical yield for znbr₂?
? g znbr₂

Explanation:

Step1: Find moles of Zn

Molar mass of Zn is \( 65.38 \, \text{g/mol} \). Moles of Zn = \( \frac{\text{mass of Zn}}{\text{molar mass of Zn}} \) = \( \frac{32.7 \, \text{g}}{65.38 \, \text{g/mol}} \approx 0.499 \, \text{mol} \)

Step2: Mole ratio of Zn to \( ZnBr_2 \)

From the reaction \( Zn + 2HBr
ightarrow ZnBr_2 + H_2 \), the mole ratio of Zn to \( ZnBr_2 \) is 1:1. So moles of \( ZnBr_2 \) = moles of Zn = \( 0.499 \, \text{mol} \)

Step3: Molar mass of \( ZnBr_2 \)

Molar mass of Zn is \( 65.38 \, \text{g/mol} \), molar mass of Br is \( 79.90 \, \text{g/mol} \). So molar mass of \( ZnBr_2 \) = \( 65.38 + 2\times79.90 = 225.18 \, \text{g/mol} \)

Step4: Mass of \( ZnBr_2 \) (theoretical yield)

Mass = moles × molar mass = \( 0.499 \, \text{mol} \times 225.18 \, \text{g/mol} \approx 112 \, \text{g} \)

Answer:

112