Magnesium metal (0.100 mol) and a volume of aqueous hydrochloric

Question

Magnesium metal (0.100 mol) and a volume of aqueous hydrochloric

acid that contains 0.500 mol of HCl are combined and react to

completion.How many liters of hydrogen gas,measured at STP, are

produced?

Balance equation first: Mg (s) HCl(aq) -> MgCl2(aq) H2 (g)

a) 2.24 L of H2

b) 4.48 L of H2

c) 5.60L of H2

d) 11.2L of H2

e) 22.4L of H2

Answer 1

Mg + 2 HCl >> MgCl2 + H2

Moles H2 = 0.100

V (H2) = 0.100 x 22.4 = 2.24 L

Answer 2

In this case you need to use the equation and the ratio of the moles. You have 0.5 mole HCl and 0.1 mole Mg. Now look at the equation, every mole of Mg reacts with one mole of HCl, so you see Mg is the 'limiting' reactant, limiting the reaction. So only 0.1 mole of HCl reacts completely with Mg. Now look at the mole ratio of Mg and H2, its 1:1, so again only 0.1 mole of H2 is produced. 1 mole of a gas= 22.4 dm^3 or 22.4L, therefore your volume is 0.1*22.4= 2.24L,

A.