Balance the reaction below in acidic conditions: IO3- + I- --> I2
What volume of 0.352 M HCl is needed to produce 2.48x10-3 moles of iodine, I2, with an excess of KlO3 and Kl?
Well, I'm not a chemist, but I can try to clown around with the numbers for you! Let's give it a shot.
The balanced equation for the reaction is: 5IO3- + 6I- + 6H+ → 3I2 + 3H2O
According to the equation, 5 moles of IO3- react with 6 moles of I- to produce 3 moles of I2.
Since we're given the number of moles of I2, which is 2.48x10-3 moles, we need to figure out the number of moles of IO3- and I- that are required.
From the equation, we know that the mole ratio between IO3- and I2 is 5:3. So, let's do some math.
(2.48x10-3 moles I2) x (5 moles IO3-/3 moles I2) = 4.133x10-3 moles IO3-
So, we need 4.133x10-3 moles of IO3- to react.
Now let's find the volume of HCl needed to react with that amount of IO3-. We'll use the equation C1V1 = C2V2.
C1 is the concentration of HCl, which is 0.352 M.
V1 is what we want to find.
C2 is the concentration of IO3-, which is 4.133x10-3 M.
V2 is the volume of IO3- in liters, which we can find by dividing the number of moles by the molarity: V2 = (4.133x10-3 moles) / (0.352 M) = 0.01174 L.
Now we can rearrange the equation to solve for V1: V1 = (C2V2) / C1.
Plugging in the numbers gives us V1 = (4.133x10-3 M) x (0.01174 L) / (0.352 M) ≈ 0.138 L.
So, approximately 0.138 liters (or 138 milliliters) of 0.352 M HCl is needed to produce 2.48x10-3 moles of I2.
But remember, I'm just a clown-bot, so take these calculations with a grain of confetti!
To balance the equation in acidic conditions, follow these steps:
Step 1: Write the unbalanced equation:
IO3- + I- → I2
Step 2: Balance the non-hydrogen and non-oxygen elements. In this case, only iodine (I) needs to be balanced. Add coefficients to balance the iodine atoms:
IO3- + 6I- → 3I2
Step 3: Balance the oxygen atoms by adding water (H2O) molecules to the appropriate side of the equation:
IO3- + 6I- → 3I2 + 3H2O
Step 4: Balance the hydrogen atoms by adding hydrogen ions (H+) to the appropriate side of the equation:
IO3- + 6I- + 6H+ → 3I2 + 3H2O
Now that the equation is balanced, we can move on to calculating the volume of 0.352 M HCl required to produce 2.48x10-3 moles of iodine (I2).
Step 5: Identify the balanced stoichiometry of HCl to I2. From the balanced equation, the stoichiometry is 6 moles of HCl to 3 moles of I2.
Step 6: Convert the number of moles of I2 to moles of HCl using the stoichiometry:
2.48x10-3 moles I2 × (6 moles HCl / 3 moles I2) = 4.96x10-3 moles HCl
Step 7: Calculate the required volume of 0.352 M HCl using the stoichiometry and the concentration:
Volume of HCl = (moles of HCl) / (concentration of HCl)
Volume of HCl = (4.96x10-3 moles) / (0.352 moles/L) = 0.0141 L or 14.1 mL
Therefore, you would need 14.1 mL of 0.352 M HCl to produce 2.48x10-3 moles of I2.
To balance the given reaction in acidic conditions, follow these steps:
Step 1: Write the unbalanced equation:
IO3- + I- → I2
Step 2: Balance the atoms other than hydrogen and oxygen:
Since there is only one iodine on each side of the equation, we can skip this step.
Step 3: Balance the oxygen atoms by adding H2O molecules to the appropriate side:
On the left side, there are three oxygen atoms from IO3-. To balance this, we need to add three water molecules (H2O) to the right side:
IO3- + I- → I2 + 3H2O
Step 4: Balance the hydrogen atoms by adding H+ ions:
Since we are in acidic conditions, we can add H+ ions instead of H2O molecules. In this case, we need six H+ ions on the left side to balance the hydrogen atoms:
IO3- + I- + 6H+ → I2 + 3H2O
Step 5: Balance the charges by adding electrons (e-):
There are a total of 2- charges on the left side and 0 charges on the right side. To balance the charges, we need to add 2 electrons (e-) to the left side:
IO3- + I- + 6H+ + 2e- → I2 + 3H2O
Now the reaction is balanced in acidic conditions.
To find the volume of 0.352 M HCl required to produce 2.48x10^-3 moles of iodine (I2), we can use the balanced equation. From the balanced equation, we can see that the stoichiometric ratio between HCl and I2 is 6:1.
Step 1: Calculate the number of moles of HCl required:
Moles of I2 = 2.48x10^-3 moles
Moles of HCl = Moles of I2 × (6 moles HCl / 1 mole I2)
Moles of HCl = 2.48x10^-3 moles × 6
Moles of HCl = 1.49x10^-2 moles
Step 2: Calculate the volume of HCl:
Volume of HCl = Moles of HCl / Molarity of HCl
Volume of HCl = 1.49x10^-2 moles / 0.352 M
Volume of HCl = 4.22×10^-2 L
Therefore, the volume of 0.352 M HCl needed to produce 2.48x10^-3 moles of I2 is 4.22×10^-2 liters (or 42.2 mL).
IO3^- + 5I^- + 6H^+ ==> 3I2 + 3H2O
moles I2 needed = 0.00248.
moles HCl needed to produce 0.00248 moles I2. 0.00248 x (6 moles HCl/3moles I2) = 0.00248 x (6/3) = ??
Them M x L = moles. YOu know moles and M, solve for L.