Al(s) + H2SO4(aq)-> Al2(SO4)3(aq) + H2(g)
Consider the unbalanced equation above. What volume of 0.490 M H2SO4 is needed to react with excess aluminum to produce 1.60 g of Al2(SO4)3? Use a molar mass with at least as many significant figures as the data given.
Well, I hope my calculations don't turn out to be a "sulfur" experience! Let's get started.
First, we need to balance the equation:
2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g)
Now, let's determine the moles of Al2(SO4)3 produced:
Molar mass of Al2(SO4)3 = 342.15 g/mol
Moles = mass / molar mass = 1.60 g / 342.15 g/mol = 0.00467 mol
According to the balanced equation, we see that 2 moles of Al react with 1 mole of Al2(SO4)3 produced. So, 0.00467 mol of Al2(SO4)3 are formed by 2 × 0.00467 = 0.00934 mol of Al.
Given that the molarity of H2SO4 is 0.490 M, we can use the molarity formula:
Molarity = moles / volume (in liters)
Rearranging the formula:
Volume = moles / molarity = 0.00934 mol / 0.490 mol/L ≈ 0.019 L
Converting liters to milliliters:
1 L = 1000 mL, so 0.019 L ≈ 19 mL
Therefore, approximately 19 mL of 0.490 M H2SO4 is needed to react with excess aluminum and produce 1.60 g of Al2(SO4)3. Always remember to stay "al-l-rounded" in your calculations!
To solve this problem, we need to follow these steps:
Step 1: Write and balance the equation:
2 Al(s) + 3 H2SO4(aq) -> Al2(SO4)3(aq) + 3 H2(g)
Step 2: Calculate the moles of Al2(SO4)3:
Given: mass of Al2(SO4)3 = 1.60 g
Molar mass of Al2(SO4)3 = 2(26.98) + 3(32.06) + 12(16.00) = 342.14 g/mol
moles = mass / molar mass
moles of Al2(SO4)3 = 1.60 g / 342.14 g/mol
Step 3: Determine the ratio of moles between H2SO4 and Al2(SO4)3:
From the balanced equation, we know that it takes 3 moles of H2SO4 to produce 1 mole of Al2(SO4)3.
Step 4: Calculate the moles of H2SO4:
moles of H2SO4 = moles of Al2(SO4)3 x (3 moles H2SO4 / 1 mole Al2(SO4)3)
Step 5: Find the volume of 0.490 M H2SO4 solution needed:
Concentration of H2SO4 = 0.490 M
Volume of H2SO4 = moles of H2SO4 / concentration
Now let's plug in the values and calculate:
moles of H2SO4 = (1.60 g / 342.14 g/mol) x (3 mol H2SO4 / 1 mol Al2(SO4)3)
moles of H2SO4 = 0.0148 mol
Volume of H2SO4 = (0.0148 mol) / (0.490 mol/L)
Volume of H2SO4 = 0.0302 L or 30.2 mL
Therefore, you would need approximately 30.2 mL of 0.490 M H2SO4 solution to react with excess aluminum and produce 1.60 g of Al2(SO4)3.
To find the volume of the 0.490 M H2SO4 needed to react with the given amount of Al2(SO4)3, we can follow these steps:
Step 1: Convert the given mass of Al2(SO4)3 to moles.
Given:
Mass of Al2(SO4)3 = 1.60 g
Molar mass of Al2(SO4)3 = ? (Let's assume it's x g/mol)
Moles of Al2(SO4)3 = Mass of Al2(SO4)3 / Molar mass of Al2(SO4)3
= 1.60 g / x g/mol
Step 2: Use the balanced equation to determine the molar ratio between H2SO4 and Al2(SO4)3.
From the balanced equation:
Al(s) + H2SO4(aq) -> Al2(SO4)3(aq) + H2(g)
The ratio between H2SO4 and Al2(SO4)3 is 1:1.
Step 3: Determine the moles of H2SO4 needed to react with the given moles of Al2(SO4)3.
Moles of H2SO4 = Moles of Al2(SO4)3 (using the molar ratio from step 2)
Step 4: Convert moles of H2SO4 to volume using the given concentration.
Given:
Concentration of H2SO4 = 0.490 M (moles per liter)
Volume of H2SO4 = Moles of H2SO4 / Concentration of H2SO4
= Moles of H2SO4 / 0.490 M
Now, let's substitute the values and calculate the volume of H2SO4 needed.
Note: Since the molar mass of Al2(SO4)3 is not provided, it is not possible to obtain the final numerical answer without that information. Please provide the molar mass of Al2(SO4)3 in order to continue the calculation and obtain the volume of H2SO4 needed.
1. Balance the equation.
2. How many moles Al2(SO4)3 do you need? mols = g/molar mass.
3. Use the equation, convert mols Al2(SO4)3 to mols H2SO4.
4. M H2SO4 = moles/L
You know M and mols, solve for L.