How many mL of 0.6 M H2SO4 solution are needed to react completely with 23.62 g of Fe2O3 in the following reaction?
Fe2O3 + 3H2SO4 → 2Fe2(SO4)3 + 3H2O
The molar mass of Fe2O3 is 159.69 g/mol.
23.62 g Fe2O3 x (1 mol Fe2O3/159.69 g Fe2O3) x (3 mol H2SO4/1 mol Fe2O3) x (0.6 L H2SO4/1 mol H2SO4) = 0.735 L H2SO4 = 735 mL H2SO4
To determine the number of mL of 0.6 M H2SO4 solution needed to react completely with 23.62 g of Fe2O3, we need to perform a stoichiometric calculation using the balanced chemical equation for the reaction.
The balanced chemical equation for the reaction between Fe2O3 and H2SO4 is as follows:
Fe2O3 + 3H2SO4 -> Fe2(SO4)3 + 3H2O
From the balanced equation, we can see that 1 mole of Fe2O3 reacts with 3 moles of H2SO4.
First, we need to calculate the number of moles of Fe2O3. To do this, we use the formula:
moles = mass / molar mass
The molar mass of Fe2O3 can be calculated by adding up the atomic masses of iron (Fe) and oxygen (O):
molar mass Fe2O3 = (2 * atomic mass Fe) + (3 * atomic mass O)
From the periodic table, we find that the atomic mass of Fe is 55.845 g/mol and the atomic mass of O is 16.00 g/mol.
molar mass Fe2O3 = (2 * 55.845 g/mol) + (3 * 16.00 g/mol)
= 111.69 g/mol + 48.00 g/mol
= 159.69 g/mol
Now, we can calculate the number of moles of Fe2O3:
moles Fe2O3 = 23.62 g / 159.69 g/mol
≈ 0.1479 mol
Since the stoichiometry of the balanced equation tells us that 1 mole of Fe2O3 reacts with 3 moles of H2SO4, we can set up the following ratio:
0.1479 mol Fe2O3 / 1 mol Fe2O3 = x mol H2SO4 / 3 mol H2SO4
Solving for x gives us:
x = (0.1479 mol Fe2O3 * 3 mol H2SO4) / 1 mol Fe2O3
≈ 0.4437 mol H2SO4
Finally, we need to convert the moles of H2SO4 to milliliters (mL) of the 0.6 M H2SO4 solution.
The molarity (M) of a solution is defined as moles per liter (mol/L).
From the given information, we have the molarity (0.6 M) and we know that x moles of H2SO4 are required.
Now we can use the formula:
Molarity = moles / volume (in liters)
To solve for the volume (V) in liters, we have:
0.6 M = 0.4437 mol H2SO4 / V
V = 0.4437 mol H2SO4 / 0.6 M
V ≈ 0.740 mL
Therefore, approximately 0.740 mL of the 0.6 M H2SO4 solution are needed to react completely with 23.62 g of Fe2O3.
To determine the amount of H2SO4 solution needed to react completely with Fe2O3, we first need to write and balance the chemical equation for the reaction:
Fe2O3 + H2SO4 -> Fe2(SO4)3 + H2O
Now, we need to calculate the moles of Fe2O3 using its molar mass. The molar mass of Fe2O3 is calculated as follows:
Fe: 55.845 g/mol * 2 = 111.69 g/mol
O: 16.00 g/mol * 3 = 48.00 g/mol
Total molar mass of Fe2O3 = 111.69 g/mol + 48.00 g/mol = 159.69 g/mol
Next, we divide the given mass of Fe2O3 by its molar mass to calculate the moles:
Moles of Fe2O3 = 23.62 g / 159.69 g/mol ≈ 0.148 moles
According to the balanced equation, the stoichiometric ratio between Fe2O3 and H2SO4 is 1:1. This means that 1 mole of Fe2O3 reacts with 1 mole of H2SO4.
Therefore, we need 0.148 moles of H2SO4 to react completely with the given amount of Fe2O3.
To calculate the volume of the H2SO4 solution, we need to use its molarity (0.6 M) and the definition of molarity, which states that:
Molarity (M) = (moles of solute) / (volume of solution in liters)
Rearranging the formula, we can solve for volume:
Volume of solution (in liters) = (moles of solute) / molarity
Now, substitute the values into the equation:
Volume of solution = 0.148 moles / 0.6 mol/L ≈ 0.247 L
Finally, we can convert the volume from liters to milliliters:
Volume of solution = 0.247 L * 1000 mL/L ≈ 247 mL
Therefore, approximately 247 mL of 0.6 M H2SO4 solution are needed to react completely with 23.62 g of Fe2O3.