How many milliliters of 0.230 M Na2S2O3 solution are needed to titrate 2.349 g of I2 according to the equation below?

I2(aq)+2S2O3^2-(aq)->S4O6^2-(aq)+2I^1-(aq)

moles I2 = grams/molar mass = ??moles.

Using the coefficients in the balanced equation, convert moles I2 to moles S2O3^-2.
??moles I2 x (2 moles S2O3^-2/1 moleI2) = xx

M S2O3 = moles/L

420 bong scopelord

To solve this problem, we need to use the concept of stoichiometry to determine the volume of the Na2S2O3 solution required to titrate the given amount of I2.

First, let's determine the number of moles of I2. We can use the formula:

moles = mass / molar mass

The molar mass of I2 is 253.808 g/mol. Plugging in the values, we have:

moles of I2 = 2.349 g / 253.808 g/mol
moles of I2 ≈ 0.00924 mol

According to the balanced equation, the stoichiometric ratio between I2 and Na2S2O3 is 1:2. This means that for every 1 mole of I2, we need 2 moles of Na2S2O3.

Therefore, we need 2 times the number of moles of Na2S2O3 compared to I2:

moles of Na2S2O3 = 2 × moles of I2
moles of Na2S2O3 = 2 × 0.00924 mol
moles of Na2S2O3 ≈ 0.01848 mol

Now, we can use the concept of molarity (M) to calculate the volume of the Na2S2O3 solution required. The formula for molarity is:

Molarity = moles / volume (in liters)

We rearrange the formula to solve for volume:

volume = moles / molarity

The molarity of the Na2S2O3 solution is given as 0.230 M. Plugging in the values, we have:

volume = 0.01848 mol / 0.230 mol/L
volume ≈ 0.0804 L

Since the volume is in liters and we need the answer in milliliters, we can convert:

volume = 0.0804 L × 1000 mL/L
volume ≈ 80.4 mL

Therefore, approximately 80.4 mL of 0.230 M Na2S2O3 solution are needed to titrate 2.349 g of I2.