Mg(s) + 2HCl(aq) --> MgCl2(aq) + H2(g)
If 1.0g of Mg is reacted with 10.0mL of 6.0M HCl, what volume of hydrogen gas will be produced at 298K & 1atm?
This is a limiting reagent problem; I know that because an amount for BOTH reactants is given. I solve these problems by working TWO simple stoichiometry problems (simple meaning not limiting reagent) for the amount of a product formed. Of course this will produce different answers but the correct value in limiting reagent problems is ALWAYS the smaller one. Here is a link to solving simple stoichiometry problems.
http://www.jiskha.com/science/chemistry/stoichiometry.html
Post your work if you get stuck.
.80
To find the volume of hydrogen gas produced, we can use the stoichiometry of the balanced equation and the ideal gas law equation.
Step 1: Convert mass of Mg to moles
Given: 1.0g of Mg
Molar mass of Mg: 24.31 g/mol
Number of moles of Mg = mass / molar mass
= 1.0g / 24.31 g/mol
≈ 0.041 moles
Step 2: Determine the limiting reactant
In this reaction, the balanced equation shows that the stoichiometric ratio between Mg and H2 is 1:1.
Since there is an excess of HCl, the limiting reactant is Mg.
Step 3: Calculate the number of moles of H2 produced
From the balanced chemical equation, we know that 1 mole of Mg produces 1 mole of H2.
Therefore, the number of moles of H2 produced is also 0.041 moles.
Step 4: Convert moles of H2 to volume using the ideal gas law
The ideal gas law equation is PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
Given:
Pressure (P) = 1 atm
Temperature (T) = 298 K
Number of moles of H2 (n) = 0.041 moles
Ideal gas constant (R) = 0.0821 L·atm/(mol·K)
Rearranging the ideal gas law equation to solve for volume:
V = nRT / P
Substituting the values:
V = (0.041 mol)(0.0821 L·atm/(mol·K))(298 K) / 1 atm
Calculating the volume:
V ≈ 0.988 L
Therefore, the volume of hydrogen gas produced is approximately 0.988 liters at 298K and 1 atm pressure.
To find the volume of hydrogen gas produced, we need to use stoichiometry and the ideal gas law.
First, let's calculate the number of moles of Mg using its molar mass. The molar mass of Mg is 24.31 g/mol, so:
Number of moles of Mg = Mass of Mg / Molar mass of Mg
= 1.0 g / 24.31 g/mol
= 0.041 mol
According to the balanced chemical equation, the stoichiometric ratio between Mg and H2 is 1:1. So, the number of moles of H2 produced will also be 0.041 mol.
Next, we can use the ideal gas law to find the volume of H2 gas. The ideal gas law equation is:
PV = nRT
Where:
P = pressure (1 atm)
V = volume (to be determined)
n = number of moles of gas (0.041 mol)
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (298 K)
Rearranging the equation to solve for V:
V = nRT / P
= (0.041 mol)(0.0821 L·atm/mol·K)(298 K) / (1 atm)
= 1.02 L
Therefore, the volume of hydrogen gas produced is 1.02 liters at 298K and 1atm.