Consider the reaction of zn metal with hydrochloric acid: zn(s) + 2hcl(aq) → zncl2(aq) + h2(g) if 7.50 g of zn is reacted with 0.225 moles of hcl in a 2.00 l container what pressure does the generated h2 exert against the container walls at 25.8 ℃?

The balanced equation for the reaction is as follows;
Zn + 2HCl --> ZnCl₂ + H₂
we need to first find which is the limiting reactant 
stoichiometry of Zn to HCl is 1:2
number of Zn moles - 7.50 g / 65.4 g/mol = 0.115 mol 
number of HCl moles - 0.225 mol 
if Zn is the limiting reactant
according to molar ratio 1:2
amount of 0.115 mol of Zn should react with - 2 x 0.115 = 0.23 mol
but only 0.225 mol of HCl present therefore HCl is the limiting reactant 
amount of products formed depends on amount of HCl present 
stoichiometry of HCl to H₂ is 2:1
number of H₂ moles formed - 0.225 / 2 = 0.113 mol

we can use the ideal gas law equation to find the pressure exerted
PV = nRT
where  
P - pressure
V - volume - 2.00 x 10⁻³ m³
n - number of moles - 0.113 mol
R - universal gas constant  - 8.314 Jmol⁻¹K⁻¹
T - temperature - 25.8 °C + 273 = 298.8 K

substituting these values in the equation 
P x 2.00 x 10⁻³ m³ = 0.113 mol x 8.314 Jmol⁻¹K⁻¹ x 298.8 K 
P = 140.4 kPa

29.1-12=17.1the answer is this one