chem caculation

Molar mass of Zn = 65 g mol-1
Molar volume of a gas at room temperature and pressure = 24 dm3

2HCl(aq) + Zn(s) → ZnCl2(g) + H2(g)
Mole ratio HCl : Zn = 2 : 1

No. of moles of HCl added = MV = 2 x (250/1000) = 0.5 mol
No. of moles of Zn added = mass/(molar mass) = 13/65 = 0.2 mol
(No. of moles of HCl added) : (No. of moles of Zn added) = 0.5 : 0.2 = 2.5 : 1
Hence, HCl is in excess, and Zn is the limiting reactant (completely reacted).

Mole ratio Zn : H2 = 1 : 1
No. of moles of Zn reacted = 0.2 mol
No. of moles of H2 formed = 0.2 mol
Volume of H2 formed = mol x (molar volume) = 0.2 x 24 = 4.8 dm3
=

2HCl+Zn---->ZnCl2+H2

no. of mole of HCl=(2.0)(250/1000) =0.5mol

no. of mole of Zn =13/65.3 =0.1991mol

Zn is limited reactant.

PV=nRT

P=(0.1991)(8.31)(273+25)/(0.1991)(24/1000)

P=103182.5Nm-2