How many liters of hydrogen gas can be made from 65.5g of magnesium at STP?
Magnesium reacts to sulfric acid according to this equation...
Mg+H2SO4-->MgSO4+H2
To determine the number of liters of hydrogen gas that can be produced from 65.5g of magnesium at STP (Standard Temperature and Pressure), we need to use stoichiometry and the given balanced equation.
Step 1: Find the molar mass of magnesium (Mg) and hydrogen (H2).
- The molar mass of magnesium (Mg) is 24.31 g/mol.
- The molar mass of hydrogen (H2) is 2.02 g/mol.
Step 2: Convert the mass of magnesium to moles.
- Moles of magnesium (Mg) = Mass of magnesium / Molar mass of magnesium
= 65.5 g / 24.31 g/mol
Step 3: Use the stoichiometric ratio from the balanced equation to convert moles of magnesium to moles of hydrogen.
- According to the balanced equation:
Mg + H2SO4 -> MgSO4 + H2
The stoichiometric ratio between magnesium (Mg) and hydrogen (H2) is 1:1.
Therefore, the moles of hydrogen (H2) produced will be equal to the moles of magnesium (Mg).
Step 4: Convert the moles of hydrogen to volume at STP.
- At STP, 1 mole of any ideal gas occupies 22.4 liters.
So, the volume of hydrogen gas produced will be equal to the moles of hydrogen (H2) produced.
Step 5: Calculate the volume of hydrogen gas.
- Volume of hydrogen gas (H2) = Moles of hydrogen gas (H2) x 22.4 L/mol
Now, let's substitute the values and calculate the volume of hydrogen gas:
Moles of magnesium (Mg) = 65.5 g / 24.31 g/mol
= 2.694 moles
Volume of hydrogen gas (H2) = 2.694 moles x 22.4 L/mol
≈ 60.29 liters
Therefore, approximately 60.29 liters of hydrogen gas can be produced from 65.5g of magnesium at STP.
mols Mg = grams/molar mass = ?
1 mol Mg = 1 mol H2 gas
mol H2 x 22.4L = ?L at STP