To calculate the mass of H2 produced, we need to use the equation and the given quantities of Mg and HCl.
First, we need to determine which reactant is limiting by finding the limiting reactant. This is done by comparing the number of moles of Mg and HCl and calculating how many moles of H2 they can produce.
Given:
Mass of Mg = 24.3 g
Mass of HCl = 75.0 g
1. Convert the masses of Mg and HCl to moles using their molar masses.
The molar mass of Mg is 24.31 g/mol, and the molar mass of HCl is approximately 36.46 g/mol.
Moles of Mg = 24.3 g / 24.31 g/mol = 1.00 mol (rounded to two decimal places)
Moles of HCl = 75.0 g / 36.46 g/mol = 2.06 mol (rounded to two decimal places)
2. From the balanced equation, we can see that 1 mol of Mg reacts with 2 mol of HCl to produce 1 mol of H2.
So, the stoichiometric ratio of HCl to H2 is 2:1.
3. Now, let's determine how many moles of H2 can be produced from the moles of HCl and Mg.
Since we have 2.06 mol of HCl, according to the stoichiometric ratio, we need 2.06 mol × (1 mol H2 / 2 mol HCl) = 1.03 mol H2 (rounded to two decimal places).
4. Finally, calculate the mass of H2 produced using its molar mass. The molar mass of H2 is approximately 2 g/mol.
Mass of H2 = 1.03 mol × 2 g/mol = 2.06 g
Therefore, the mass of H2 produced is 2.06 grams.