We drop 50.5 grams of magnesium into 572 mL of a 3 M HCl solution. What is the maximum volume of dry hydrogen that could be produced by this reaction at STP?
To find the maximum volume of hydrogen gas produced by the reaction, we first need to balance the chemical equation for the reaction between magnesium (Mg) and hydrochloric acid (HCl). The balanced equation is as follows:
Mg + 2HCl → MgCl2 + H2
From the balanced equation, we can see that one mole of magnesium reacts with two moles of hydrochloric acid to produce one mole of hydrogen gas.
Step 1: Convert grams of magnesium to moles
To find the number of moles of magnesium, we divide the given mass of magnesium by its molar mass. The molar mass of magnesium (Mg) is 24.31 g/mol.
Moles of Mg = Mass of Mg / Molar mass of Mg
Moles of Mg = 50.5 g / 24.31 g/mol
Step 2: Determine the limiting reactant
Next, we compare the number of moles of magnesium (Mg) to the number of moles of hydrochloric acid (HCl) to determine the limiting reactant. We can use the molarity and volume of the HCl solution to find the number of moles of HCl using the formula:
Moles of HCl = Molarity × Volume (in liters)
Given that the molarity (M) of the HCl solution is 3 M and the volume is 572 mL, we first need to convert the volume to liters:
Volume (in liters) = 572 mL ÷ 1000 mL/L
Moles of HCl = 3 M × (572 mL ÷ 1000 mL/L)
Since the stoichiometric ratio between Mg and HCl is 1:2, we multiply the moles of HCl by the ratio (2) to find the equivalent moles of Mg:
Equivalent moles of Mg = Moles of HCl × 2
Step 3: Calculate the maximum volume of dry hydrogen gas at STP
According to Avogadro's Law, 1 mole of any gas occupies 22.4 liters at standard temperature and pressure (STP). Since the balanced equation shows that the ratio of Mg to H2 is 1:1, the moles of Mg will also be equal to the moles of H2.
Hence, the maximum volume of dry hydrogen gas produced can be calculated as follows:
Volume of H2 = Moles of H2 × 22.4 L/mol
Now we can put all the values together to find the answer.