what volume of hydrogen gas, measured at STP, will be produced when 16.0g of magnesium reacts completely with an excess of hydrochloric acid
To determine the volume of hydrogen gas produced when magnesium reacts with hydrochloric acid, we need to follow these steps:
1. Write and balance the chemical equation for the reaction:
Mg + 2HCl → MgCl2 + H2
2. Calculate the number of moles of magnesium (Mg) used:
- Find the molar mass of magnesium (Mg) from the periodic table. It is approximately 24.31 g/mol.
- Use the formula: Moles = Mass / Molar mass
Moles of Mg = 16.0 g / 24.31 g/mol = 0.658 mol (rounded to 3 decimal places)
3. Use the stoichiometry of the balanced chemical equation to determine the moles of hydrogen gas (H2) produced:
From the balanced equation, 1 mole of magnesium produces 1 mole of hydrogen gas.
Moles of H2 = 0.658 mol (Since it is a 1:1 ratio)
4. Use the ideal gas law to calculate the volume of the gas produced at STP (Standard Temperature and Pressure):
- The molar volume of a gas at STP is 22.4 L/mol.
- Use the formula: Volume = Moles x Molar volume
Volume of H2 = 0.658 mol x 22.4 L/mol = 14.72 L (rounded to 3 decimal places)
Therefore, approximately 14.72 L of hydrogen gas will be produced at STP when 16.0 g of magnesium reacts completely with an excess of hydrochloric acid.
Mg + 2HCl = MgCl2 + H2
So, each mole of Mg produces 1 mole of H2
So, how many moles of Mg is 16g?
1 mole of H2 occupies 22.4L at STP.