Question

Use the equation Zn + 2HCl -> ZnCl₂ + H₂

_{If you react 71.24 g Zn with 91.53 g HCl, what mass of zinc chloride is produced? And what is the limiting reactant in this situation? Show all calculations.}

Answer 1

Zn + 2HCl -> ZnCl2 + H2

65.3 g 2 x 36.5 g = 73 g

Given that 71.24 g Zn reacted with 91.53 g HCl.

Hence,

Moles of Zn = mass / molar mass = 71.24 g / 65.3 g/mol = 1.09 mol

Moles of HCl = mass / molar mass = 91.53 g / 73 g/mol = 1.25 mol

Here, Zn present in lesser no of moles. Therefore, Zn is the limiting reagent and yield is calculated based on Zn.

Calculation of mass of zinc chloride produced

Zn + 2HCl -> ZnCl2 + H2

65.3 g   136.3 g

71.24 g ?

? = (71.24 g Zn/ 65.3 g Zn ) x 136.3 g ZnCl2

= 148.7 g ZnCl2

mass of ZnCl2 = 148.7 g ZnCl2

  Therefore, 148.7 g ZnCl2 wil be produced.