We drop 17.3 grams of magnesium into
435 mL of a 2 M HCl solution. What is
the maximum volume of dry hydrogen that
could be produced by this reaction at STP?
Mg + 2HCl ==> H2 + MgCl2
This is a limiting reagent problem. We know that because amounts are given for BOTH reactants.
mols Mg = grams/molar mass = ?
mols HCl = M x L = ?
Using the coefficients in the balanced equation, convert mols Mg to mols H2.
Do the same for mols HCl to mols H2.
It is likely that the two values obtained for mols H2 gas will not agree which means one of them is not right. The correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
For the smaller value, at STP 1 mol H2 occupies 22.4L; therefore, volume H2 = mols H2 x 22.4 L/mol = ? L.
To find the maximum volume of dry hydrogen produced by the reaction, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
To solve this problem, we need to follow these steps:
Step 1: Determine the moles of each reactant:
Given that the mass of magnesium is 17.3 grams, and the molar mass of magnesium is 24.31 g/mol, we can calculate the moles of magnesium using the formula:
moles of magnesium = mass / molar mass
moles of magnesium = 17.3 g / 24.31 g/mol
Similarly, for the hydrochloric acid (HCl), we are given the concentration as 2 M and the volume as 435 mL. We can first convert the volume to liters by dividing by 1000:
volume of HCl = 435 mL / 1000 mL/L
Then we can calculate the moles of HCl using the formula:
moles of HCl = volume * concentration
Step 2: Determine the stoichiometric ratio:
From the balanced chemical equation between magnesium (Mg) and hydrogen gas (H2), we know that:
1 mole of Mg reacts with 2 moles of HCl to produce 1 mole of H2
Step 3: Identify the limiting reactant:
To determine the limiting reactant, compare the mole ratios calculated in Step 1 with the stoichiometric ratio in Step 2. The reactant that provides fewer moles in comparison to the stoichiometry is the limiting reactant.
Step 4: Calculate the maximum volume of dry hydrogen:
Since we now know the limiting reactant, we can use the stoichiometric ratio to find the moles of hydrogen gas produced. Then, we can use the ideal gas law equation to calculate the volume of hydrogen gas at STP.
The ideal gas law equation is:
PV = nRT
P = pressure at STP (standard temperature and pressure) = 1 atm
V = volume of hydrogen gas at STP (what we need to find)
n = moles of hydrogen gas (calculated from stoichiometric ratio and limiting reactant)
R = ideal gas constant = 0.0821 L•atm/(mol•K)
T = temperature at STP = 273 K
By rearranging the ideal gas law equation, we can solve for V:
V = (n * R * T) / P
Substituting the values, we can now calculate the maximum volume of dry hydrogen gas that could be produced.
To determine the maximum volume of dry hydrogen produced, we need to calculate the number of moles of magnesium reacted and then use the stoichiometry of the balanced chemical equation to find the number of moles of hydrogen produced. Finally, we can use the ideal gas law to calculate the volume of hydrogen at STP.
First, let's calculate the number of moles of magnesium (Mg) reacted:
Given:
Mass of Mg = 17.3 g
Atomic mass of Mg = 24.31 g/mol
Number of moles of Mg = Mass of Mg / Atomic mass of Mg
= 17.3 g / 24.31 g/mol
Next, let's use the stoichiometry to find the number of moles of hydrogen (H2) produced:
The balanced chemical equation for the reaction between Mg and HCl is:
Mg + 2HCl → MgCl2 + H2
From the equation, we see that 1 mole of Mg reacts with 2 moles of HCl to produce 1 mole of H2.
Number of moles of H2 = Number of moles of Mg
Now, let's calculate the volume of hydrogen produced at STP using the ideal gas law:
At STP (Standard Temperature and Pressure), the temperature is 273.15 K and the pressure is 1 atm.
The ideal gas law is given by:
PV = nRT
Where:
P = Pressure (in atm)
V = Volume (in liters)
n = Number of moles
R = Gas constant (0.0821 L.atm/mol.K)
T = Temperature (in Kelvin)
Since we want to calculate the volume, we can rearrange the equation as follows:
V = nRT / P
Substituting the values:
n = Number of moles of H2
R = 0.0821 L.atm/mol.K
T = 273.15 K
P = 1 atm
Finally, we can calculate the volume of dry hydrogen produced:
V = (Number of moles of H2) * (0.0821 L.atm/mol.K) * (273.15 K) / (1 atm)