What is the minimum amount of 6.7M H2SO4 necessary to produce 27.9g of H2 (g) according to the following reaction? 2Al(s)+3H2SO4(aq)→Al2(SO4)3(aq)+3H2(g)
I don't know how to solve it, and I don't understand the explanation
given online.
You worked all of those stoichiometry problems (even the limiting reagent problem) beautifully that I checked last night. This is the same thing with just a slight twist at the end.
2Al(s)+3H2SO4(aq)→Al2(SO4)3(aq)+3H2(g)
mols H2 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols H2 to mols H2SO4.
Then M H2SO4 = mols H2SO4/L H2SO4. You know mols from above and M from the problem, solve for L H2SO4. Convert to mL if you wish.
To calculate the minimum amount of 6.7M H2SO4 necessary to produce 27.9g of H2 (g), we need to follow these steps:
Step 1: Convert the given mass of H2 to moles.
We can use the molar mass of H2 to convert the mass to moles.
The molar mass of H2 is 2 g/mol.
To calculate the moles of H2, divide the mass by the molar mass:
27.9g H2 / 2 g/mol = 13.95 mol H2
Step 2: Use the balanced chemical equation to determine the stoichiometry.
According to the balanced chemical equation:
2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g)
we can see that 3 moles of H2SO4 are required to produce 3 moles of H2.
So, the molar ratio between H2SO4 and H2 is 3:3, or 1:1.
Step 3: Calculate the moles of H2SO4 needed.
Since the molar ratio between H2SO4 and H2 is 1:1, the moles of H2SO4 needed will also be 13.95 mol.
Step 4: Calculate the volume of the 6.7M H2SO4 solution needed.
The concentration of the H2SO4 solution is given as 6.7M, which means there are 6.7 moles of H2SO4 in 1 liter (1000mL) of solution.
To calculate the volume of the solution needed, divide the moles of H2SO4 needed by the molarity of the solution:
Volume of H2SO4 solution = (moles of H2SO4 needed) / (molarity of H2SO4 solution)
Volume of H2SO4 solution = 13.95 mol / 6.7 mol/L ≈ 2.08 L
Therefore, the minimum amount of 6.7M H2SO4 necessary to produce 27.9g of H2 (g) is approximately 2.08 liters of the solution.
To solve this problem, we need to use stoichiometry, which is a way to relate the quantities of substances in a chemical reaction.
First, we need to determine the molar mass of H2SO4 (sulfuric acid). The molar mass of H2SO4 is 98.09 g/mol.
Next, we need to determine the number of moles of H2 that is produced in the reaction. We have 27.9 g of H2, so we can use the molar mass of H2 (2 g/mol) to calculate the number of moles:
moles of H2 = mass / molar mass = 27.9 g / 2 g/mol = 13.95 mol
Now, we can use the balanced equation to determine the stoichiometric ratio between H2SO4 and H2. From the equation, we can see that 3 moles of H2 are produced from 3 moles of H2SO4:
2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g)
Therefore, the ratio is 3 moles of H2SO4 : 3 moles of H2 or 1 mole of H2SO4 : 1 mole of H2.
Finally, we can use the stoichiometric ratio to calculate the amount of H2SO4 needed to produce 13.95 moles of H2:
moles of H2SO4 = moles of H2
Therefore, the minimum amount of H2SO4 necessary to produce 27.9 g of H2 is 13.95 moles.
But we need to convert the moles to grams, so we multiply by the molar mass of H2SO4:
mass of H2SO4 = moles of H2SO4 * molar mass of H2SO4
mass of H2SO4 = 13.95 mol * 98.09 g/mol
mass of H2SO4 = 1,367.52 g
Therefore, the minimum amount of 6.7M H2SO4 necessary to produce 27.9g of H2 is 1,367.52 grams.
I hope this explanation helps clarify the steps involved in solving this problem using stoichiometry. Let me know if you have any other questions!