35.04 mL of .237M I2 solution is required to titrate a sample containing As2O3. Calculate the mass of As2O3 (197.8 g/mol) in the sample.
As2O3 + 5(H2O) + 2I2 --> 2(H3AsO4) + 4HI
2007-11-05 07:45:10 · 4 answers · asked by Peter 2 in Science & Mathematics ➔ Chemistry
As2O3 + 5(H2O) + 2I2 --> 2(H3AsO4) + 4HI
Moles I2
35.04 ml * 0.237 moles/1000ml = 0.0083 moles
each moles I2 will require 1/2 moles of As2O3
moles As2O3 require = 0.0083/2 = 0.00415 moles
As2O3 MW = 197.8 g/mole
0.00415 moles As2O3 = 0.00415 mole * 197.8 g/mole =
= 0.821 g As2O3 in the sample
2007-11-05 07:57:39 · answer #1 · answered by Dr Dave P 7 · 0⤊ 0⤋
The mole ratio of I2 to As2O3 is 2:1. The number of moles of I2 in the 35.04mL is 0.237*35.04/1000 which equals 0.0083. Therefore half as much moles of As2O3 are present which equals 0.00415 moles. Multiplying this by the molecular mass gives (0.00415x 197.8) 0.821grams.
2007-11-05 16:00:02 · answer #2 · answered by haile d 3 · 0⤊ 0⤋
moles(I2) = 35.04 x 0.237 / 1000 = 8.30448 x 10 ^-3 mol
From the reaction equation 2 moles iodine react with one mole of arsenic oxide.
So 8.30448 X 10^-3 mol.(I2) is equivalent to 2
hence 4.15224 x 10^-3 mol.(As2O3) is equivalent to 1.
moles = mass/Mr
mass = moles x Mr
mass = 4.15224 x 10^-3 x 197.8 = 0.8213 grams (4 dp.)
2007-11-05 16:07:33 · answer #3 · answered by lenpol7 7 · 0⤊ 0⤋
First solve for your moles of I2
(.237 mol/L)(.03504)=.00830448 mol I2
the ratio from the balanced equation is 1/2 so
(1mol As2O3/2 mol I2)(.00830448mol I2)= .00415224 mol As2O3
Multiple by the molar mass to find how many grams
(.00415224mol)(197.8g/mol)=.821g As2O3
2007-11-05 15:58:14 · answer #4 · answered by Apium 2 · 0⤊ 0⤋