How many liters of hydrogen gas is produced from 3.712 g of magnesium with 104.2ml of 1.385 mol/L HCL (aq) at SATP? Please help me!! I need to figure this question out for a quiz.

The correct answer is 1.61 L.

Explanation:

Based on the given information, 3.712 grams of Mg reacts with 104.2 ml of 1.385 mol per L HCl at SATP, there is a need to find the amount of hydrogen gas produced in liters.  

The chemical reaction taking place in the given case is,  

Mg + 2HCl = MgCl2 + H2

The reacting moles of each reactants is,  

Moles of Mg = 3.712 g/24.305 g = 0.153 moles

Moles of HCl = 1.385 mol/L * 0.1042 L = 0.144 moles

From the reaction it is clear that Mg and HCl are present in 1:2 molar ratio. Therefore, 0.153 moles of Mg can completely react with 0.306 moles of HCl. However, the moles of HCl obtained in the given case is only 0.144 moles, thus, HCl is a limiting reactant.  

Now the moles of hydrogen produced is,  

n = 0.144 moles of HCl * (1 mole H2/2 mol HCl) = 0.072 moles

Finally to find the liters of hydrogen gas produced, the ideal gas equation is used, that is, PV = nRT

At STAP, the value of T is 273 K and pressure is 1 atm, the value of R is 0.082 atm.L/mol.L. Now putting the values we get,  

PV = nRT

V = nRT/P

V = 0.072 mol * 0.082 atm.L/mol.L*273/ 1 atm

V = 1.61 L

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