An excess of mg (s) is added to 100. ml of 0.400 m hcl. at 0c and 1 atm pressure, what volume of h 2 (g) can be obtained?

a. 44.8 ml

b. 224 ml

c. 448 ml

d. 896 ml

The balanced equation for the reaction between Mg and HCl is as follows

Mg + 2HCl - - > MgCl₂ + H₂

stoichiometry of HCl to H₂ is 2:1

number of HCl moles reacted - 0.400 mol/L x 0.100 L = 0.04 mol of HCl

since Mg is in excess HCl is the limiting reactant

number of H₂ moles formed - 0.04/2 = 0.02 mol of H₂

we can use ideal gas law equation to find the volume of H₂

PV = nRT

where

P - pressure - 1 atm x 101 325 Pa/atm = 101 325 Pa

V - volume

n - number of moles - 0.02 mol

R - universal gas constant - 8.314 Jmol⁻¹K⁻¹

T - temperature in Kelvin - 0 °C + 273 = 273 K

substituting these values in the equation

101 325 Pa x V = 0.02 mol x 8.314 Jmol⁻¹K⁻¹ x 273 K

V = 448 x 10⁻⁶ m³

V = 448 mL

therefore answer is

c. 448 mL