If 16.01 grams of sulfuric acid are reacted with a large amount of aluminum metal, what volume of hydrogen gas would be produced at STP?
What is the balanced equation?
3H2SO4+2Al>>Al2(SO4)3 + 6H2
how many moles of sulfuric acid is in 16.01 grams?
To determine the balanced equation for the reaction between sulfuric acid and aluminum, we first need to identify the chemical formulas of sulfuric acid and aluminum and then write the equation in a balanced form.
The chemical formula for sulfuric acid is H2SO4, and aluminum is represented by Al. The reaction between sulfuric acid and aluminum can be represented as follows:
2 Al + 3 H2SO4 → Al2(SO4)3 + 3 H2
This equation shows that 2 moles of aluminum react with 3 moles of sulfuric acid to produce 1 mole of aluminum sulfate (Al2(SO4)3) and 3 moles of hydrogen gas (H2).
Now let's move on to calculating the volume of hydrogen gas produced at STP (standard temperature and pressure). At STP, 1 mole of any gas occupies 22.4 liters of volume.
To find the volume of hydrogen gas produced, we need to determine the number of moles of hydrogen gas first. We can use the concept of stoichiometry to find the moles of hydrogen gas.
Given that 16.01 grams of sulfuric acid are reacted, we can calculate the number of moles of sulfuric acid by dividing the mass by its molar mass.
Molar mass of H2SO4 = (2 * 1.008 g/mol) + (32.07 g/mol) + (4 * 16.00 g/mol) = 98.09 g/mol
Moles of H2SO4 = Mass / Molar mass = 16.01 g / 98.09 g/mol = 0.163 mol
According to the balanced equation, the molar ratio between H2SO4 and H2 is 3:3 or 1:1. Therefore, the number of moles of hydrogen gas produced is also 0.163 mol.
Now we can calculate the volume of hydrogen gas at STP using the relationship between moles and volume at STP (22.4 L/mol).
Volume of H2 gas = Number of moles * Volume at STP = 0.163 mol * 22.4 L/mol ≈ 3.6488 L ≈ 3.65 L
Therefore, approximately 3.65 liters of hydrogen gas would be produced at STP when 16.01 grams of sulfuric acid reacts with aluminum.