Using I2 (aq)+ 2S2O3=> S4O6 (aq)+ 2I(aq) calculate the number of moles of molecular iodine. Given that there are 0.004328482 Moles of Na2S2O3.
To calculate the number of moles of molecular iodine (I2) produced in the reaction, we need to use the molar ratio between Na2S2O3 (sodium thiosulfate) and I2.
According to the balanced equation:
2I2(aq) + Na2S2O3(aq) -> Na2S4O6(aq) + 2I(aq)
The molar ratio between Na2S2O3 and I2 is 2:2, which means that for every 2 moles of Na2S2O3, 2 moles of I2 are produced.
Given that there are 0.004328482 moles of Na2S2O3, we can use this information to calculate the number of moles of I2.
Number of moles of Na2S2O3 = 0.004328482 moles
Using the molar ratio (2:2), we can say that:
0.004328482 moles of Na2S2O3 = 0.004328482 moles of I2
Therefore, the number of moles of molecular iodine (I2) produced in the reaction is 0.004328482 moles.
To calculate the number of moles of molecular iodine (I2), we can use the balanced chemical equation and the stoichiometry of the reaction.
The balanced chemical equation for the reaction is:
I2 (aq) + 2S2O3 (aq) → S4O6 (aq) + 2I (aq)
According to the equation, for every 1 mole of I2, 2 moles of Na2S2O3 (sodium thiosulfate) are consumed.
Given that there are 0.004328482 moles of Na2S2O3, we can use the stoichiometric ratio between Na2S2O3 and I2 to find the number of moles of I2.
Stoichiometric ratio: 1 mole I2 : 2 moles Na2S2O3
Therefore, to calculate the number of moles of I2, we divide the moles of Na2S2O3 by the stoichiometric coefficient that corresponds to I2:
Number of moles of I2 = 0.004328482 moles Na2S2O3 / 2
Number of moles of I2 = 0.002164241 moles
Therefore, there are 0.002164241 moles of molecular iodine (I2) in the given reaction.