What mass of Cu(IO3)2 can be formed from 0.690g of CuSO4-5H2O?
Work from CuSO4-5H2O + 2IO3^- => Cu(IO3)2 + SO4^2-
Convert 0.690g to moles => 0.0027mole
Since the equation molar ratio is 1:1 for CuSO4-5H2O & Cu(IO3)2 then the moles of Cu(IO3)2 => 0.0027mole.
Convert 0.0027mole Cu(IO3)2 to grams => (0.0027mole)(413.53g/mole)
=>1.14gms Cu(IO3)2
Well, let's do some chemistry calculations! To find the mass of Cu(IO3)2 that can be formed, we need to consider the molar ratios of the reactants and products.
First, let's look at the molar mass of CuSO4-5H2O:
- CuSO4 = 159.6 g/mol
- H2O = 18.0 g/mol (5 molecules)
So, the molar mass of CuSO4-5H2O is 159.6 + (18.0 * 5) = 249.6 g/mol.
Now let's focus on the reaction equation:
CuSO4-5H2O + 2 NaIO3 → Cu(IO3)2 + Na2SO4 + 5 H2O
Here, we can see that the molar ratio between CuSO4-5H2O and Cu(IO3)2 is 1:1. So, the molar mass of CuSO4-5H2O is also equal to the molar mass of Cu(IO3)2.
Therefore, the 0.690 g of CuSO4-5H2O will yield 0.690 g of Cu(IO3)2. Just like magic! Well, not really magic, but you know what I mean.
To determine the mass of Cu(IO3)2 formed from the given mass of CuSO4-5H2O, we need to use stoichiometry.
1. Write the balanced chemical equation for the reaction:
CuSO4 · 5H2O + 2 NaIO3 → Cu(IO3)2 + Na2SO4 + 5 H2O
2. Find the molar masses of the compounds involved:
CuSO4 · 5H2O:
Cu: 63.55 g/mol
S: 32.07 g/mol
O: 16.00 g/mol (x 4 for S)
H: 1.01 g/mol (x 10 for H)
Total molar mass: 249.69 g/mol
Cu(IO3)2:
Cu: 63.55 g/mol
I: 126.90 g/mol (x 2 for I)
O: 16.00 g/mol (x 6 for O)
Total molar mass: 425.45 g/mol
3. Use stoichiometry to find the molar ratio between CuSO4·5H2O and Cu(IO3)2:
1 mole CuSO4 · 5H2O produces 1 mole Cu(IO3)2
4. Calculate the number of moles of CuSO4 · 5H2O:
Moles = mass / molar mass = 0.690 g / 249.69 g/mol = 0.00276 mol
5. Use the stoichiometric ratio to find the number of moles of Cu(IO3)2:
Moles of Cu(IO3)2 = Moles of CuSO4 · 5H2O = 0.00276 mol
6. Calculate the mass of Cu(IO3)2:
Mass = moles × molar mass = 0.00276 mol × 425.45 g/mol = 1.17 g
Therefore, the mass of Cu(IO3)2 that can be formed from 0.690 g of CuSO4-5H2O is approximately 1.17 grams.
To determine the mass of Cu(IO3)2 that can be formed from CuSO4·5H2O, you need to use stoichiometry and the balanced chemical equation for the reaction.
1. Start by writing the balanced equation for the reaction between CuSO4·5H2O and IO3- ions:
CuSO4·5H2O + 2 IO3- -> Cu(IO3)2 + H2SO4 + 5 H2O
2. Calculate the molar mass of CuSO4·5H2O:
Cu = 63.55 g/mol
S = 32.07 g/mol
O = 16.00 g/mol (4 oxygens)
H = 1.008 g/mol (10 hydrogens)
Add the molar masses together:
CuSO4·5H2O = (63.55 + 32.07 + (16.00 x 4) + (1.008 x 10)) g/mol = 249.68 g/mol
3. Use the molar mass to convert grams of CuSO4·5H2O to moles:
Moles = Mass / Molar Mass
Moles = 0.690 g / 249.68 g/mol = 0.00276 mol
4. Use stoichiometry to find the moles of Cu(IO3)2 formed:
According to the balanced equation, the mole ratio between CuSO4·5H2O and Cu(IO3)2 is 1:1. This means that for every 1 mole of CuSO4·5H2O, 1 mole of Cu(IO3)2 is formed.
Therefore, the number of moles of Cu(IO3)2 formed is also 0.00276 mol.
5. Convert moles of Cu(IO3)2 to grams:
Moles = Mass / Molar Mass
Mass = Moles x Molar Mass
Mass = 0.00276 mol x (63.55 + (16.00 x 6)) g/mol = 0.512 g
So, the mass of Cu(IO3)2 that can be formed from 0.690g of CuSO4·5H2O is 0.512 g.