Question

What volume (in mL!!!) of 2.25 M HCl is required to react with 2.03 g of zinc (65.41 g/mol) according to the following reaction? Zn(s) + 2 HCl (aq) → ZnCl2(aq) + H2(g)

Answer 1

number of mol of Zn = (given mass of Zn)/(molar mass of Zn)
= 2.03/65.41
= 0.031 mol

reaction taking place is
Zn(s) + 2 HCl (aq) --> ZnCl2(aq) + H2(g)

according to reaction
1 mol of Zn required 2 mol of HCl
0.031 mol of Zn required (2*0.031) mol of HCl
0.031 mol of Zn required 0.062 mol of HCl
so,
number of mol of HCl required = 0.062 mol

use,
volume of HCl required = (number of mol of HCl required)/(molarity)
= 0.062/2.25
= 0.0276 L
= 27.6 mL

Answer: 27.6 mL