Oxalic acid (H2C2O4) can be used to remove rust according to the following balanced equation:
Fe2O3 + 6H2C2O4 --> 2Fe(C2O4)33- +3H2O + 6H+
Calculate the number of grams of rust that can be removed by 5.00 x 102 ml of a 0.122M solution of oxalic acid.
My first thought is since when is rust IronIII oxide?
Ok, how many moles of oxalic acid?
moles=.500l*.122=you do it.
Now, moles of IronIIIoxide=moleacid *6)
Then, convert moles of ironIII oxide to grams.
6.6 grams of rust
To calculate the number of grams of rust that can be removed by a given volume of oxalic acid solution, you need to follow these steps:
Step 1: Convert the volume of the oxalic acid solution from milliliters (mL) to liters (L).
To convert milliliters to liters, you divide the given volume by 1000:
5.00 x 10^2 mL ÷ 1000 = 0.500 L
Step 2: Calculate the number of moles of oxalic acid in the solution using the molarity (M) and volume (V) of the solution.
The equation is: moles = Molarity x Volume
moles = 0.122 M x 0.500 L = 0.061 mol
Step 3: Use the balanced equation to determine the ratio of moles of rust (Fe2O3) to moles of oxalic acid (H2C2O4).
From the balanced equation, the stoichiometric ratio is:
1 mol Fe2O3 : 6 mol H2C2O4
Step 4: Convert moles of oxalic acid to moles of rust using the stoichiometric ratio.
moles of rust = 0.061 mol H2C2O4 x (1 mol Fe2O3 / 6 mol H2C2O4) = 0.0102 mol Fe2O3
Step 5: Calculate the molar mass of Fe2O3.
The molar mass of Fe2O3 is the sum of the atomic masses of iron (Fe) and oxygen (O), taking into account the subscripts in the formula:
2(55.845 g/mol Fe) + 3(16.00 g/mol O) = 159.69 g/mol
Step 6: Calculate the mass of rust using the moles of rust and the molar mass of Fe2O3.
mass of rust = moles of rust x molar mass of Fe2O3
mass of rust = 0.0102 mol Fe2O3 x 159.69 g/mol = 1.63 g
Therefore, 5.00 x 10^2 mL of a 0.122 M solution of oxalic acid can remove approximately 1.63 grams of rust.