Magnesium metal (0.100 mol) and a volume...

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 litres of hydrogen gas, measured at STP are produced?

Answer 1

The first step would be to write the balance equation for the reaction, which is...

Mg(s) + 2HCl(aq) --> MgCl₂ (aq)+ H₂ (aq)

Now you must determine whether Mg or HCl is the limiting reactant. By dividing the moles of each substance that you are given by its coefficient in the balanced equation, the smallest result will come from the limiting reactant. In this case, it is Mg, because 0.100/1 (= 0.100) is less than 0.500/2 (= 0.250). Therefore, the amount of Mg will determine the amount of H₂ gas that can be produced. According to the balanced equation, 1 mol Mg will produce 1 mol of H₂ gas, therefore 0.100 mol of Mg will produced 0.100 mol H₂ . Finally, you plug in all your known values into the ideal gas law and solve for the volume of H₂ ...

PV=nRT

(1atm)(V) = (0.100mol)(0.082057338 L * atm / mol * K)(273.15K)

V = 2.24L

Answer 2

The balanced reaction follows:

2Mg_(s) + 2HCl_(aq) → H_2(g) + 2MgCl_(s)

Since the stoichiometric coefficients are equal for both reactants, the Mg is the limiting reagent because the amount (0.100 mol) is lesser than the amount of HCl (0.500 mol).

So the amount of H₂ produced is given by:

0.100 mol Mg x (1 mol H₂) / (2 mol Mg) = 0.050 mol H₂

Solve for the amount of H₂ in liters, given STP conditions (1 atm and 273.15 K), the gas constant R = 0.0821 L•atm/(mol•K), and the Ideal Gas Law (PV = nRT):

PV = nRT

V = nRT / P

V = (0.050 mol) x (0.0821 L•atm/(mol•K)) x (273.15 K) / (1 atm) = 1.12 L

** A helpful note on problems such as this is to pay attention to units. Knowing that we are solving for a value in volume units (L) makes selecting the correct equation (Ideal Gas Law) and correct value for the gas constant (R) much easier.**