Final answer:
To determine the number of moles of I2 consumed by 2.80 mL of 0.15 M Na2S2O3, first calculate the number of moles of Na2S2O3 using the concentration and volume. Then use the balanced chemical equation to determine the stoichiometric ratio between Na2S2O3 and I2. Finally, multiply the moles of Na2S2O3 by the stoichiometric ratio to find the moles of I2 consumed.
Step-by-step explanation:
To determine how many moles of I2 can be consumed by 2.80 mL of 0.15 M Na2S2O3, we can use the concentration and volume of Na2S2O3 to calculate the number of moles of Na2S2O3 present. Then, using the balanced chemical equation for the reaction between Na2S2O3 and I2, we can determine the stoichiometric ratio between Na2S2O3 and I2. Finally, we can multiply the number of moles of Na2S2O3 by the stoichiometric ratio to find the number of moles of I2 consumed.
Using the formula C = n/V, where C is the concentration in moles per liter, n is the number of moles, and V is the volume in liters, we can rearrange the formula to solve for n: n = C * V. Plugging in the values, we have n = 0.15 M * 0.00280 L = 0.00042 moles of Na2S2O3.
The balanced equation for the reaction between Na2S2O3 and I2 is:
2 Na2S2O3 + I2 → 2 NaI + Na2SO4
According to the balanced equation, 2 moles of Na2S2O3 react with 1 mole of I2. Therefore, if we have 0.00042 moles of Na2S2O3, we can calculate the number of moles of I2 consumed by multiplying by the stoichiometric ratio: 0.00042 moles Na2S2O3 * (1 mole I2 / 2 moles Na2S2O3) = 0.00021 moles of I2 consumed.