2. If you used 0.500 g of Mg and an excess of HCL, how much hydrogen gas should be produced at STP?
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To find out how much hydrogen gas will be produced at STP (Standard Temperature and Pressure) when reacting 0.500 g of magnesium (Mg) with an excess of hydrochloric acid (HCl), we need to use stoichiometry.
Step 1: Write the balanced chemical equation for the reaction between magnesium and hydrochloric acid:
Mg + 2HCl -> MgCl2 + H2
Step 2: Calculate the moles of Mg using its molar mass:
Molar mass of Mg = 24.31 g/mol
Moles of Mg = mass of Mg / molar mass of Mg
= 0.500 g / 24.31 g/mol
Step 3: Determine the stoichiometric ratio between Mg and H2 from the balanced equation:
From the balanced equation, we can see that 1 mole of Mg reacts to produce 1 mole of H2.
Step 4: Calculate the moles of H2 produced:
Moles of H2 = Moles of Mg (since the stoichiometric ratio is 1:1)
Step 5: Use the ideal gas law to calculate the volume of H2 at STP:
The ideal gas law equation is PV = nRT, where
P = pressure (at STP, it is 1 atm),
V = volume,
n = moles of gas,
R = ideal gas constant (0.0821 L·atm/mol·K),
T = temperature (at STP, it is 273.15 K).
Rearranging the equation, we get V = nRT / P.
Substituting the values into the equation:
V = (moles of H2) * (R) * (T) / (P)
V = (moles of H2) * (0.0821 L·atm/mol·K) * (273.15 K) / (1 atm)
Step 6: Calculate the volume of H2 gas at STP:
Volume of H2 gas = (moles of H2) * (22.4 L/mol)
(At STP, 1 mole of any gas occupies 22.4 L)
Now, let's perform the calculations:
Step 2 (recap): Moles of Mg = 0.500 g / 24.31 g/mol
Step 4: Moles of H2 = Moles of Mg = 0.500 g / 24.31 g/mol
Step 6: Volume of H2 gas = (moles of H2) * (22.4 L/mol)