We drop 37.4 grams of magnesium into
461 mL of a 3 M HCl solution. What is
the maximum volume of dry hydrogen that
could be produced by this reaction at STP?
Mg(s) + 2HCl(aq) ! MgCl2(aq) + H2(g)
Answer in units of liters
To find the maximum volume of dry hydrogen gas produced at STP, we need to calculate the number of moles of magnesium (Mg) used and then use the stoichiometry of the balanced chemical equation to determine the moles of hydrogen gas (H2) produced. Finally, we can use the ideal gas law to convert the moles of hydrogen gas to the volume at STP.
Step 1: Calculate the moles of magnesium (Mg) used
To calculate the moles of magnesium, we need to use the molar mass of magnesium (24.31 g/mol):
moles of Mg = mass of Mg / molar mass of Mg
moles of Mg = 37.4 g / 24.31 g/mol
moles of Mg ≈ 1.537 mol
Step 2: Calculate the moles of hydrogen gas (H2) produced
According to the balanced chemical equation, 1 mole of magnesium reacts with 2 moles of HCl to produce 1 mole of H2:
moles of H2 = moles of Mg x (moles of H2 / moles of Mg)
moles of H2 = 1.537 mol x (1 mol H2 / 1 mol Mg)
moles of H2 ≈ 1.537 mol
Step 3: Convert moles of hydrogen gas to volume at STP
We can use the ideal gas law equation PV = nRT to convert the moles of hydrogen gas to volume at STP. At STP (Standard Temperature and Pressure), the temperature (T) is 273.15 K (0 °C) and the pressure (P) is 1 atm:
V = (n x R x T) / P
V = (1.537 mol x 0.0821 L·atm/mol·K x 273.15 K) / 1 atm
V ≈ 33.76 L
Therefore, the maximum volume of dry hydrogen gas that could be produced is approximately 33.76 liters.
To find the maximum volume of dry hydrogen that could be produced by the given reaction, we can use stoichiometry. Let's break down the steps to find the answer:
Step 1: Calculate the number of moles of magnesium (Mg).
To find the number of moles of Mg, we'll use the equation:
moles of Mg = mass of Mg / molar mass of Mg
The molar mass of Mg is 24.31 g/mol. Plugging in the values:
moles of Mg = 37.4 g / 24.31 g/mol = 1.537 mol
Step 2: Determine the limiting reactant.
To determine the limiting reactant, we'll compare the number of moles of Mg with the number of moles of HCl. According to the balanced chemical equation, for every 1 mole of Mg, 2 moles of HCl are required.
Since the ratio of moles between Mg and HCl is 1:2, we have:
moles of HCl = 2 * moles of Mg
moles of HCl = 2 * 1.537 mol = 3.074 mol
Step 3: Calculate the volume of hydrogen gas (H2) at STP.
According to the balanced chemical equation, 1 mole of Mg produces 1 mole of H2 gas. Therefore, the moles of H2 gas produced would be equal to the moles of Mg used.
Using the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature, we can solve for the volume (V) of H2 gas.
At STP (standard temperature and pressure), the values are:
P = 1 atm
T = 273.15 K (0 °C)
R = 0.0821 L·atm/(mol·K)
V = nRT / P
V = 1.537 mol * 0.0821 L·atm/(mol·K) * 273.15 K / 1 atm
V ≈ 33.6 L
Therefore, the maximum volume of dry hydrogen gas that could be produced by this reaction at STP is approximately 33.6 liters.