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.
I know i need to multiply the # of moles*6*160 (which is the molecular mass)
.5z*.122=
0.061 moles of oxalic acid.
so is it .061*6*160= 585.6
is that right i'm pretty sure its not
You error is here
0.500 L x 0.122 M = 0.061 moles and NOT 0.610 moles
I got 1.63 g agreed
oh okay i got it
thanks
The oxalic acid solution contains
0.500 x 0.122 moles of oxalic acid.
As 6 moles of oxalic acid react with 1 moles of rust then 0.500 x 0.122 moles will react with
0.500 x 0.122 x 1/6 moles of rust
the MM for rust is 159.6882 g mole^-1
so the mass of rust is
0.500 x 0.122 x 1/6 x 159.6882 g
(3 sig figs are appropriate int he answer)
Your answer leaves me with 1.63 so is this wrong moles oxalic acid = 0.500 L x 0.122 M=0.610
the ratio between Fe2O3 and H2C2O4 is 1 : 6
moles Fe2O3 = 0.610/6=0.102
mass Fe2O3 = 0.102 mol x 159.69 g/mol=16.3 g
It left me with 16.3
To calculate the number of grams of rust that can be removed by the given amount of oxalic acid solution, we need to use stoichiometry and the balanced equation.
First, let's calculate the number of moles of oxalic acid in the solution:
Volume of Solution = 5.00 x 10^2 mL = 5.00 x 10^-1 L
Concentration of Oxalic Acid = 0.122 M
Number of Moles of Oxalic Acid = Concentration x Volume
= 0.122 mol/L x 5.00 x 10^-1 L
= 0.061 mol
Next, we use the balanced equation to determine the stoichiometric ratio between oxalic acid and rust:
Fe2O3 + 6H2C2O4 --> 2Fe(C2O4)33- + 3H2O + 6H+
According to the equation, 6 moles of oxalic acid react with 1 mole of rust (Fe2O3).
Therefore, the number of moles of rust that can be removed is:
Number of moles of rust = 0.061 mol x (1 mol Fe2O3 / 6 mol H2C2O4)
= 0.0102 mol
Finally, we can calculate the mass of rust using the molar mass of Fe2O3:
Molar Mass of Fe2O3 = (2 x Atomic Mass of Fe) + (3 x Atomic Mass of O)
= (2 x 55.845) + (3 x 15.999)
= 159.691 g/mol
Mass of Rust = Number of moles of rust x Molar mass of Fe2O3
= 0.0102 mol x 159.691 g/mol
= 1.63 grams
Therefore, 5.00 x 10^2 ml of a 0.122 M solution of oxalic acid can remove approximately 1.63 grams of rust.