What is the molarity of ZnCl2 that forms when 20.0 g of zinc completely reacts with CuCl2 according to the following reaction? Assume a final volume of 260 mL . Zn(s)+CuCl2(aq)→ZnCl2(aq)+Cu(s)

Question

mass of Zn = 20.0 g

molar mass of Zn = 65.4 g/mol

number of moles of Zn = mass / molar mass= 20.0/65.4= 0.306 mol Zn = moles of ZnCl2 formed  

(since the stoichiometric factor is 1:1)

volume = 260 mL = 0.260 L

0.306mol Zn/ 0.260 L = 1.18 M

Answer 1

20.0 g Zn. x. 1 mol Zn x. 1 mol ZnCl₂ = 0.306 mol ZnCl₂<- First convert mass of Zn to moles, then using mole

65.38 g Zn 1 mol Zn ratio from the balanced equation ( 1 to 1)

M = mol/L. = 0.306 mol / 0.260 L = 1.2 M. <- Next use the molarity equation. Be sure to convert volume to liters. Remember significant figures! 260 mL is 2 sf!