How many L of hydrogen gas at 1.11atm and 25.0oC will be produced from 16.7g of magnesium reacting with an excess amount of hydrochloric acid, HCl? Mg + 2HCl → MgCl2 + H2

Answer 1

First, write the correctly balanced equation for the reaction taking place:

Mg(s) + 2HCl(aq) ==> MgCl₂(aq) + H2(g)

Next, since HCl is present in excess, there is no need to find the limiting reactant, as it will be Mg.

So, find moles of Mg, then moles H₂ and finally the volume of H₂.

Moles Mg = 16.7 g Mg x 1 mol Mg/24.305 g = 0.6871 moles Mg

Moles H₂ produced = 0.6871 moles Mg x 1 mol H₂/1 mol Mg = 0.6871 moles H₂

To find volume of H2, we use the ideal gas law; PV = nRT and solve for V (volume):

V = nRT/P

n = moles H2 = 0.6871 moles

R = gas constant = 0.08206 Latm/Kmol

T = temperature in K = 25.0 + 273 = 298K

P = pressure = 1.11 atm

V = (0.6871)(0.08206)(298)/1.11 = 15.1 L (to 3 significant figures)