Hydrogen gas can be produced by reacting aluminum with sulfuric acid. How many moles of sulfuric acid are needed to completely react with 15.0 mol of aluminum?
2Al(s) + 3H2SO4(aq) Al2(SO4)3(aq) + 3H2(g)
the equation says that for every 2 moles of Al, you need 3 moles of H2SO4.
So, you will need 3/2 * 15 = 22.5 moles of H2SO4
Well, it seems like you're trying to put aluminum and sulfuric acid on a blind date and expecting them to produce some hydrogen gas! Let's calculate the number of moles of sulfuric acid that are needed for this magical reaction.
According to the balanced equation, you need 3 moles of sulfuric acid for every 2 moles of aluminum. Since we have 15.0 moles of aluminum, we can set up a simple proportion:
2 moles Al / 3 moles H2SO4 = 15.0 moles Al / x moles H2SO4
Cross-multiplying and solving for x, we get:
2x = 3 * 15.0
2x = 45.0
x = 45.0 / 2
x = 22.5
So, you would need 22.5 moles of sulfuric acid to completely react with 15.0 moles of aluminum. Just make sure to invite the right amount of sulfuric acid to the party, or your reaction might not go with a bang!
From the balanced equation, it is clear that the mole ratio between aluminum (Al) and sulfuric acid (H2SO4) is 2:3. This means that for every 2 moles of aluminum, we need 3 moles of sulfuric acid.
Given that we have 15.0 mol of aluminum, we can use this mole ratio to find the number of moles of sulfuric acid needed.
15.0 mol Al * (3 mol H2SO4 / 2 mol Al) = 22.5 mol H2SO4
Therefore, 22.5 moles of sulfuric acid are needed to completely react with 15.0 moles of aluminum.
To find out how many moles of sulfuric acid are needed to completely react with 15.0 moles of aluminum, we can use the mole ratio from the balanced chemical equation.
In the balanced equation:
2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g)
From the equation, we see that 2 moles of aluminum react with 3 moles of sulfuric acid to produce 3 moles of hydrogen gas.
This tells us that the mole ratio of aluminum to sulfuric acid is 2:3.
So, if we have 15.0 moles of aluminum, we can set up a proportion:
2 moles of aluminum / 3 moles of sulfuric acid = 15.0 moles of aluminum / x moles of sulfuric acid
Solving for x, the number of moles of sulfuric acid needed, we get:
x = (15.0 moles of aluminum * 3 moles of sulfuric acid) / 2 moles of aluminum
x = 22.5 moles of sulfuric acid
Therefore, 22.5 moles of sulfuric acid are needed to completely react with 15.0 moles of aluminum.