After writing a balanced equation for the reaction between magnesium and hydrochloric acid, calculate the mass of magnesium needed to produce 90ml of hydrogen gas @ STP conditions.
Here is a worked example of a stoichiometry problem. Just follow the steps. Remember, L/22.4 will convert volume of H2 to moles H2.
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To calculate the mass of magnesium needed, we need to follow these steps:
Step 1: Write the balanced chemical equation for the reaction between magnesium (Mg) and hydrochloric acid (HCl).
The balanced equation is:
Mg + 2HCl → MgCl2 + H2
Step 2: Determine the molar ratio between magnesium and hydrogen gas.
From the balanced equation, we can see that 1 mole of magnesium (Mg) reacts to produce 1 mole of hydrogen gas (H2).
Step 3: Convert the volume of hydrogen gas to moles using the ideal gas law at standard temperature and pressure (STP) conditions.
According to the ideal gas law, 1 mole of any gas occupies 22.4 liters at STP conditions. So, to convert 90 ml of hydrogen gas to liters, we divide by 1000 to get 0.09 liters. Then, we divide 0.09 liters by 22.4 liters/mole to obtain the moles of hydrogen gas.
0.09 L ÷ 22.4 L/mol = 0.004 mole of H2
Step 4: Use the molar ratio to find the moles of magnesium needed.
Since the molar ratio between magnesium and hydrogen gas is 1:1, we also have 0.004 mole of magnesium.
Step 5: Calculate the molar mass of magnesium.
The molar mass of magnesium (Mg) is 24.31 g/mol.
Step 6: Determine the mass of magnesium using the moles and molar mass.
Mass of magnesium = Moles of magnesium × Molar mass of magnesium
Mass of magnesium = 0.004 mol × 24.31 g/mol = 0.09724 g
Therefore, approximately 0.097 g of magnesium is needed to produce 90 ml of hydrogen gas at STP conditions.