What volume of dry hydrogen would be produced under standard consitions when 0.250g of magnesium reacts with excess HCl?
For this sort of problem always start with a balanced equation,
Mg + HCl -> H2 + MgCl2
which you need to balance.
molar mass for Mg is 24.3 g mole^-1
number of moles (M) of Mg is
M= 0.260 g / 24.3 g mole^-1
Then use the equation to find the number of moles of H2.
Under standard conditios one mole of any gas occupires 22.4 litres
Hence find the volume of gas.
To determine the volume of dry hydrogen produced when magnesium reacts with excess hydrochloric acid, we need to follow these steps:
Step 1: Write and balance the chemical equation for the reaction.
The balanced chemical equation for the reaction between magnesium (Mg) and hydrochloric acid (HCl) is as follows:
Mg + 2HCl → MgCl2 + H2
Step 2: Calculate the number of moles of magnesium.
Given that the mass of magnesium (Mg) is 0.250g, we need to convert this mass into moles. To do this, we use the molecular weight (atomic mass) of magnesium, which is approximately 24.31 g/mol.
Number of moles (n) = mass (m) / molecular weight (M)
n = 0.250g / 24.31 g/mol
n = 0.0103 mol
Step 3: Determine the stoichiometric ratio.
From the balanced chemical equation, we can see that 1 mole of Mg reacts with 1 mole of H2. Therefore, the molar ratio between Mg and H2 is 1:1.
Step 4: Calculate the number of moles of hydrogen produced.
Since the molar ratio between Mg and H2 is 1:1, the number of moles of hydrogen produced will also be 0.0103 mol.
Step 5: Apply Avogadro's Law to find the volume of hydrogen at standard conditions.
According to Avogadro's Law, equal volumes of gases, at the same temperature and pressure, contain an equal number of molecules or moles.
At standard temperature and pressure (STP), the conditions are defined as 0 degrees Celsius (273.15 K) and 1 atmosphere (1 atm) pressure.
1 mole of gas occupies 22.4 liters at STP.
Therefore, the volume of dry hydrogen produced when 0.250g of magnesium reacts with excess HCl under standard conditions is:
Volume (V) = number of moles (n) × molar volume at STP
V = 0.0103 mol × 22.4 liters/mol
V ≈ 0.231 liters or 231 milliliters
So, approximately 0.231 liters (231 mL) of dry hydrogen gas would be produced under standard conditions.