What is the minimum amount of 6.0M H2SO4 necessary to produce 19.0g of H2(g) according to the reaction?
2Al(s)+3H2SO4(aq)→Al2(SO4)3(aq)+3H2(g)
To determine the minimum amount of 6.0M H2SO4 necessary to produce 19.0g of H2(g) according to the given reaction, we can use stoichiometry.
Stoichiometry is a tool used in chemistry to relate the amounts of substances involved in a chemical reaction. It allows us to convert between the masses of different substances using their respective mole ratios.
First, let's calculate the molar mass of H2 (hydrogen gas):
H2: 2(1.01 g/mol) = 2.02 g/mol
Next, we need to determine the moles of H2(g) needed. To do this, we'll use the molar mass of H2 and the given mass of H2(g):
moles of H2 = mass of H2 / molar mass of H2
moles of H2 = 19.0 g / 2.02 g/mol ≈ 9.41 mol
Now, let's write the balanced equation:
2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g)
According to the balanced equation, we need 3 moles of H2SO4 to produce 3 moles of H2.
Finally, we can calculate the volume (in liters) of 6.0M H2SO4 required to react with the calculated moles of H2SO4:
moles of H2SO4 = 3/3 = 1 mol
Molarity (M) is defined as the number of moles of solute per liter of solution. Since we know the molarity of H2SO4 is 6.0M, we can rearrange the equation to solve for volume:
Molarity (M) = moles / volume (L)
Therefore,
volume (L) = moles / Molarity (M)
volume (L) = 1 mol / 6.0 mol/L = 0.167 L
Thus, the minimum amount of 6.0M H2SO4 necessary to produce 19.0g of H2(g) is approximately 0.167 liters.
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 and M, solve for L.