Create an image portraying a science experiment scene to visualize the conversion reaction described. On one side, depict a small heap of bluish crystals, symbolizing 0.2750 g of CuSO4 · 5H2O. On the other side, illustrate a container with white powder, representing KIO3, waiting to be quantified. In the middle, sketch a beautifully crafted copper-colored crystal structure, signifying Cu(IO3)2. Use lab elements like weighing scales, chemical flasks, and safety goggles to set the background.

What mass of KIO3 is needed to convert the copper in 0.2750 g of CuSO4 · 5H2O to

Cu(IO3)2?

Well, that's a chemistry question that requires some calculations! But I'm here to bring some laughter into the equation. So, let me try to clown around with the numbers before we actually calculate anything!

Why did the copper go to the circus? Because it wanted to be transformed into Copper(I)!

Now, let's get to the calculations. First, we need to find the molar mass of Cu(IO3)2:

Cu(IO3)2 = (1 Cu + 2 I + 6 O) x (63.55 g/mol + 2 x 126.9 g/mol + 6 x 16.00 g/mol)
= 531.2 g/mol

Next, we'll convert the mass of CuSO4 · 5H2O to moles:

0.2750 g CuSO4 · 5H2O x (1 mol CuSO4 · 5H2O / 249.69 g CuSO4 · 5H2O) = 0.00110 mol CuSO4 · 5H2O

Since the molar ratio between CuSO4 · 5H2O and Cu(IO3)2 is 1:1, we need the same number of moles of Cu(IO3)2. So:

0.00110 mol Cu(IO3)2 x (531.2 g Cu(IO3)2 / 1 mol Cu(IO3)2) = 0.584 g KIO3

Therefore, you would need approximately 0.584 grams of KIO3 to convert the copper in 0.2750 grams of CuSO4 · 5H2O to Cu(IO3)2.

To calculate the mass of KIO3 needed to convert the copper in CuSO4 · 5H2O to Cu(IO3)2, we will follow these steps:

Step 1: Write balanced chemical equations for the reactions that occur.

The reaction between CuSO4 · 5H2O and KIO3 can be represented as:
2 CuSO4 · 5H2O + 6 KIO3 → Cu(IO3)2 + 3 K2SO4 + 10 H2O

Step 2: Calculate the molar mass of CuSO4 · 5H2O.
The molar mass of CuSO4 · 5H2O is the sum of the molar masses of its individual elements.

Cu: 1 * atomic mass of Cu
S: 1 * atomic mass of S
O: 4 * atomic mass of O
H: 2 * atomic mass of H
O: 5 * atomic mass of O
Total molar mass = Cu + S + 4O + 2H + 5O

Step 3: Calculate the number of moles of CuSO4 · 5H2O.
To calculate the number of moles, we use the given mass and the molar mass of CuSO4 · 5H2O.
Number of moles = Given mass / Molar mass

Step 4: Use stoichiometry to calculate the moles of KIO3 needed.
From the balanced chemical equation, the mole ratio between CuSO4 · 5H2O and KIO3 is 2:6.
Moles of KIO3 = Moles of CuSO4 · 5H2O * (6/2)

Step 5: Calculate the mass of KIO3 needed.
Mass = Moles * Molar mass

Now let's perform the calculations:

Step 2:
Molar mass of Cu = 63.55 g/mol
Molar mass of S = 32.07 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of K = 39.10 g/mol
Molar mass of I = 126.90 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of CuSO4 · 5H2O = (Cu + S + 4O + 2H + 10O) g/mol = (63.55 + 32.07 + (4 * 16.00) + (2 * 1.01) + (10 * 16.00)) g/mol

Step 3:
Number of moles of CuSO4 · 5H2O = 0.2750 g / Molar mass of CuSO4 · 5H2O

Step 4:
Moles of KIO3 = Moles of CuSO4 · 5H2O * (6/2)

Step 5:
Mass of KIO3 = Moles of KIO3 * Molar mass of KIO3

By following these steps, you should be able to calculate the mass of KIO3 needed to convert the copper in 0.2750 g of CuSO4 · 5H2O to Cu(IO3)2.

To find the mass of KIO3 needed to convert the copper in CuSO4 · 5H2O to Cu(IO3)2, we need to use stoichiometry. Stoichiometry is a way to calculate the quantities of reactants and products in a chemical reaction.

First, let's write the balanced chemical equation for the reaction:

2 CuSO4 · 5H2O + 4 KIO3 → Cu(IO3)2 + K2SO4 + 5 H2O

From the balanced equation, we can see that the molar ratio between CuSO4 · 5H2O and KIO3 is 2:4 or 1:2.

Now, let's calculate the molar mass of CuSO4 · 5H2O and use it to determine the number of moles of CuSO4 · 5H2O.

The molar mass of CuSO4 · 5H2O can be calculated as:

(1 mol Cu) + (1 mol S) + (4 mol O) + (10 mol H) + (4 mol O) = 249.7 g/mol

To find the number of moles of CuSO4 · 5H2O, we divide the given mass (0.2750 g) by the molar mass:

0.2750 g / 249.7 g/mol = 0.00110 mol

Since the molar ratio between CuSO4 · 5H2O and KIO3 is 1:2, the number of moles of KIO3 required is twice the number of moles of CuSO4 · 5H2O, which is 0.00220 mol.

Finally, to find the mass of KIO3 required, we multiply the number of moles by the molar mass of KIO3:

0.00220 mol * (1 mol K) + (1 mol I) + (3 mol O) = 214.0 g/mol = 0.471 g

Therefore, the mass of KIO3 needed to convert the copper in 0.2750 g of CuSO4 · 5H2O to Cu(IO3)2 is 0.471 g.

Long way and short way. mm stands for molar mass. I've rounded molar masses and not watched the number of significant figures.

Short way:
0.2750 g CuSO4.5H2O x (2*mm KIO32/mm CuSO4.5H2O) = ?estimated 0.47 g.
Long way:
CuSO4.5H2O + 2KIO3 ==> Cu(IO3)2 + K2SO4 + 5H2O
mols CuSO4.5H2O = grams/molar mass = 0.2750/249.5 = 0.001102
0.001102 mols CuSO4.5H2O x (2 mols KIO3/1 mol CuSO4.5H2O) = 0.002204 mols KIO3.
grams KIO3 = mols x molar mass = 0.002204 x 214 = about 0.47 g